Calculus 2 - Infinite Series

Contents: Integral Infinite Series Sequence Series First Order Differential Equation Reference: Paul's Online Notes: Series And Sequences

Sequence

Definition
A sequence {a_n} is a function whose domain is the set of positive integers. The functional values a₁, a₂, ..., a_n, ... are the terms of the sequence. The term a_n is the n-th term of the sequence.

Limit of a Sequence
A sequence {a_n} converges and has the limit L, or $\lim_{n\to \infty}a_n=L$ if for every Ɛ > 0, there exists a positive integer N such that |a_n - L| < Ɛ whenever n > N.

Squeeze Theorem for Sequence If there exists some integer N such that an ≤ bn ≤ cn for all n ≥ N and $\lim_{n\to \infty}a_n= \lim_{n\to \infty}c_n = L$, then $\lim_{n\to \infty}b_n= L$.

Theorem I If $\lim_{n\to \infty}a_n= L$ and the function f is continuous at L, then $\lim_{n\to \infty}f(a_n)= f (\lim_{n\to \infty}a_n) = f(L)$

Theorem II If $\lim_{n\to \infty} |a_n|= 0$, then $\lim_{n\to \infty} a_n= 0$.

Theorem III The sequence ${r^n}$ converges if $-1 < r \leq 1$.

  Example 5.1 - Convergence of Sequence

Determine whether the given sequence converges or diverges. If it converges, find the limit.
$\begin{align*} &\text{A. } \left \{ \frac{\sin{\sqrt{n}}}{\sqrt{n}} \right \}\\ &\text{B. } \left \{ \frac{e^n}{n^2} \right \}\\ &\text{C. } \left \{ \left | \frac{(-1)^n}{n} \right | \right \}\\ &\text{D. } \left \{ 1 + 2(-1)^n \right \} \\ &\text{E. } \left \{ \tanh{x} \right \} \\ \end{align*}$

Solution:

1. Range of sin function:
   $\begin{align*} -1 \leq &\sin{\sqrt{n}} \leq 1\\ -\frac{1}{\sqrt{n}} \leq & \frac{\sin{\sqrt{n}}}{\sqrt{n}} \leq \frac{1}{\sqrt{n}}\\ \lim_{n\to \infty} -\frac{1}{\sqrt{n}} & \leq \lim_{n\to \infty} \frac{\sin{\sqrt{n}}}{\sqrt{n}} \leq \lim_{n\to \infty} \frac{1}{\sqrt{n}} \end{align*}$
   
   Through Squeeze Theorem, we can see that the sequence converges to 0.


2. Since (Using L'Hôpital's rule):
   $\begin{align*} &\lim_{n\to \infty} \frac{e^n}{n^2} \\ &= \lim_{n\to \infty} \frac{e^n}{2n} \\ &= \lim_{n\to \infty} \frac{e^n}{2} \\ &= \infty \end{align*}$
   
   The sequence diverges.


3. Since
   $\begin{align*} & \lim_{n\to \infty} \left |\frac{(-1)^n}{n} \right| \\ &= \lim_{n\to \infty} \frac{1}{n}\\ &= 0 \end{align*}$
   
   The sequence converges to 0.


4. Since
   $\begin{align*} & \lim_{n\to \infty} \left ( 1 + 2(-1)^n \right ) \\ &= \lim_{n\to \infty} 1 + \lim_{n\to \infty} 2(-1)^n \\ &= 1 + 2 \lim_{n\to \infty} (-1)^n \\ \end{align*}$
   
   Through Theorem III, we can see that the entire sequence diverges since r = -1.


5. Since
   $\begin{align*} & \lim_{n\to \infty} \tanh{x} \\ &= \lim_{n\to \infty} \frac{\sinh{x}}{\cosh{x}} \\ &= \lim_{n\to \infty} \frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}} \\ &= \lim_{n\to \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}} \\ &= \lim_{n\to \infty} \frac{1 - \frac{1}{e^{2x}} }{1 + \frac{1}{e^{2x}}} \\ &= 1 \end{align*}$
   
   The sequence converges to 1.

Series

Definition
An expression of the form: $a_1 + a_2 +a_3 + ... + a_n + ...$ is called a series, or an infinite series. The series can be denoted by: $\sum_{n=1}^{\infty}a_n$

Convergence of Infinite Series
Give an infinite series:
$\sum_{n=1}^{\infty}a_n = a_1 + a_2 +a_3 + ... + a_n + ...$

The n-th partial sum of that series is S_n:
$S_n= \sum_{k=1}^{n}a_k = a_1 + a_2 +a_3 + ... + a_n$

If the sequence of partial sum {S_n} converges to the number S, or if $\lim_{n\to \infty} S_n = S$, then the series Σa_n converges to S, or written:
$\sum_{n=1}^{\infty}a_n = a_1 + a_2 +a_3 + ... + a_n + ... = S$

Otherwise, if {S_n} diverges, then the series Σa_n diverges.

Geometric Series
A series in the form:
$\sum_{n=1}^{\infty}ar^{n-1} = a+ar+ar^2+...+ar^{n-1}+...,\;a\neq0$
is called a geometric series with common ratio r.

A geometric series with |r| < 1 converges to $\frac{a}{1-r}$. Otherwise, it diverges.

P Series
A series in the form:
$\sum_{n=1}^{\infty} \frac{1}{n^p}$
will be convergent if p > 1. Otherwise, divergent.

  Example 5.2 - Convergence of Series

Some of the examples below show the method to determine whether a series converges or diverges from the sum of the series.

Determine whether the given series converges or diverges. If it converges, find the limit. (Series D and E are Geometric Series)
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\ &\text{B. } \sum_{n=2}^{\infty} \left ( \frac{1}{n^2+3n+2} \right ) \\ &\text{C. } \sum_{n=1}^{\infty} \left ( \frac{7}{3n} \right ) \\ &\text{D. } \frac{5}{3} - \frac{5}{9} + \frac{5}{27} - \frac{5}{81} + ... \\ &\text{E. } \sum_{n=0}^{\infty} \left ( 2^n 3^{-n+1} \right ) \\ \end{align*}$

Solution:

1. Since
   $\begin{align*} &\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right )\\ &=\left (1- \frac{1}{2} \right ) + \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + ... + \left (\frac{1}{n} - \frac{1}{n+1} \right )\\ &=1 - \frac{1}{n+1} \end{align*}$
   
   We have just obtained the sum of the series. Next:
   $\lim_{n\to \infty} \left ( 1 - \frac{1}{n+1} \right ) = 1$
   
   So, that series converges to 1.


2. For this problem, recall Partial Fractions:
   $\begin{align*} \frac{1}{n^2+3n+2}&=\frac{1}{(n+2)(n+1)}\\ &=\frac{A}{n+2}+\frac{B}{n+1}\\ 1 &= A(n+1) + B(n+2)\\ &= (A+B)n + A + 2B\\ \\ A=-B\\ 1 &= A + 2B\\ A&=-1\\ B&=1 \end{align*}$
   
   Now that we have found the alternate form of the series, we can try finding its sum:
   $\begin{align*} &\sum_{n=0}^{\infty} \left ( \frac{1}{n+1} - \frac{1}{n+2} \right )\\ &=1 - \frac{1}{n+2} \end{align*}$
   
   The way we find the sum of the series above is exactly the same as the previous problem. The same terms are simplified, so that only the first and last term remains. This series and the series in the previous problem are Telescoping Series. Lastly:
   $\lim_{n\to \infty} \left ( 1 - \frac{1}{n+2} \right ) = 1$
   
   So, that series also converges to 1.


3. We can rewrite the series as:
   $\sum_{n=0}^{\infty} \frac{7}{3n} = \frac{7}{3} \sum_{n=0}^{\infty} \frac{1}{n}\\ $
   
   The series above is a P series with r=1. So, that series is divergent.


4. This is a geometric series with ratio $-\frac{1}{3}$. A geometric series will be convergent if: -1 < r < 1. So, we can see that this series is convergent...
   $S = \frac{a}{1-r} = \frac{\frac{5}{3}}{1+\frac{1}{3}} = \frac{5}{4}$
   ...to $\frac{5}{4}$.


5. We usually recognize geometric series through the multiplying ratios. Though this may not look like a geometric series at first glance, let us manipulate the form:
   $\begin{align*} &\sum_{n=0}^{\infty} \left ( 2^n 3^{-n+1} \right )\\ &= \sum_{n=0}^{\infty} \left ( 2^n 3^{1-n} \right ) \\ &= \sum_{n=0}^{\infty} \left ( 2^n \left (\frac{3}{3^n} \right ) \right ) \\ &= \sum_{n=0}^{\infty} 3 \left ( \left (\frac{2}{3} \right )^n \right ) \end{align*}$
   
   Now, we know that the ratio of this geometric series is $\frac{2}{3}$. A geometric series will be convergent if: -1 < r < 1. So, we can see that this series is convergent...
   $S = \frac{a}{1-r} = \frac{3}{1-\frac{2}{3}} = 9$
   ...to 9.

Tests to Determine Convergence/Divergence of Series

Divergence Test If $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n\to \infty}a_n = 0$ If $\lim_{n\to \infty}a_n$ does not exist or $\lim_{n\to \infty}a_n \neq 0$, then $\sum_{n=1}^{\infty} a_n$ diverges.

  Example 5.3 - Divergence Test

Some of the examples below show the method to determine whether a series converges or diverges through the Divergence Test.

Keep in mind that the Divergence Test can only be used to PROVE that a series is divergent.



Show that the given series are divergent, using the Divergence Test:
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \frac{4n^2-n^3}{7-3n^3}\\ &\text{B. } \sum_{n=1}^{\infty} \frac{1}{2+3^{-n}}\\ &\text{C. } \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...\\ &\text{D. } \sum_{n=1}^{\infty} \cos{\frac{1}{n^3}}\\ &\text{E. } \sum_{n=1}^{\infty} \frac{n!}{n}\\ \end{align*}$

Solution:

1. 
   $\begin{align*} &\lim_{n\to \infty} a_n\\ &= \lim_{n\to \infty} \frac{4n^2-n^3}{7-3n^3}\\ &= \lim_{n\to \infty} \frac{\frac{4}{n}-1}{\frac{7}{n^3}-3}\\ &= \frac{1}{3} \end{align*}$
   
   Since that limit ≠ 0, it is shown that the series is divergent.


2. 
   $\begin{align*} &\lim_{n\to \infty} a_n\\ &= \lim_{n\to \infty} \frac{1}{2+3^{-n}}\\ &= \lim_{n\to \infty} \frac{1}{2+\frac{1}{3^n}}\\ &= \frac{1}{2} \end{align*}$
   
   Since that limit ≠ 0, it is shown that the series is divergent.


3. The series above can be rewritten as:
   $\sum_{n=1}^{\infty} \frac{n}{n+1}$
   
   $\begin{align*} &\lim_{n\to \infty} a_n\\ &= \lim_{n\to \infty} \frac{n}{n+1}\\ &= 1 \end{align*}$
   
   Since that limit ≠ 0, it is shown that the series is divergent.


4. 
   $\begin{align*} &\lim_{n\to \infty} a_n\\ &= \lim_{n\to \infty} \cos{\frac{1}{n^3}} \\ &= \cos { \lim_{n\to \infty} \frac{1}{n^3}} \\ &= \cos{0}\\ &= 1 \end{align*}$
   
   Since that limit ≠ 0, it is shown that the series is divergent.


5. 
   $\begin{align*} &\lim_{n\to \infty} a_n\\ &= \lim_{n\to \infty} \frac{n!}{n}\\ &= \lim_{n\to \infty} (n-1)!\\ &= \lim_{n\to \infty} (n-1)(n-2)(n-3)...(2)(1)\\ &= \infty \end{align*}$
   
   Since that limit ≠ 0, it is shown that the series is divergent.

Comparison Test If $\lim_{n\to \infty}\frac{a_n}{b_n} > 0$, then $\sum a_n$ and $\sum b_n$ are both divergent, or convergent.

  Example 5.4 - Comparison Test

Determine whether the given sequence converges or diverges using the Comparison Test.
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \frac{3n^2+1}{2n^5+n+2}\\ &\text{B. } \sum_{n=1}^{\infty} \frac{\sqrt{2n^2+4n+1}}{n^3+9}\\ &\text{C. } \sum_{n=1}^{\infty} \frac{n}{2^n-1}\\ &\text{D. } \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+1}\\ &\text{E. } \sum_{n=1}^{\infty} \frac{n^2}{n^3+\cos^2{n}}\\ \end{align*}$

Solution:

1. When n approaches a really large number, we can assume that the function behaves like $\frac{n^2}{n^5} = \frac{1}{n^3}$.
   
   We will use that as b_n, and the initial function as a_n.We can see that $b_n = \frac{1}{n^3}$ is convergent.
   $\begin{align*} &\lim_{n\to \infty} \frac{a_n}{b_n}\\ &= \lim_{n\to \infty} \frac{3n^2+1}{2n^5+n+2} \times n^3\\ &= \lim_{n\to \infty} \frac{3n^5+n^3}{2n^5+n+2}\\ &= \frac{3}{2} \end{align*}$
   
   By the definition of Comparison Test, we can determine that the series is convergent.


2. When n approaches a really large number, we can assume that the function behaves like $\frac{\sqrt{n^2}}{n^3} = \frac{1}{n^2}$.
   
   We will use that as b_n, and the initial function as a_n. We can see that $b_n = \frac{1}{n^2}$ is convergent.
   $\begin{align*} &\lim_{n\to \infty} \frac{a_n}{b_n}\\ &= \lim_{n\to \infty} \frac{\sqrt{2n^2+4n+1}}{n^3+9} \times n^2\\ &= \lim_{n\to \infty} \frac{\sqrt{2n^6+4n^5+n^4}}{n^3+9}\\ &= \lim_{n\to \infty} \frac{\sqrt{2+\frac{4}{n}+\frac{1}{n^2}}}{1+\frac{9}{n^3}}\\ &= \sqrt{2} \end{align*}$
   
   By the definition of Comparison Test, we can determine that the series is convergent.


3. When n approaches a really large number, we can assume that the function behaves like $\frac{n}{2^n}$.
   
   We will use that as b_n, and the initial function as a_n. We can see that $b_n = \frac{n}{2^n}$ is convergent.
   $\begin{align*} &\lim_{n\to \infty} \frac{a_n}{b_n}\\ &= \lim_{n\to \infty} \frac{n}{2^n - 1} \times \frac{2^n}{n}\\ &= \lim_{n\to \infty} \frac{\sqrt{n(2^n)}}{n(2^n)-n}\\ &= \lim_{n\to \infty} \frac{1}{1 - \frac{1}{2^n}}\\ &= 1 \end{align*}$
   
   By the definition of Comparison Test, we can determine that the series is convergent.


4. When n approaches a really large number, we can assume that the function behaves like $\frac{1}{\sqrt{n}}=\frac{1}{n^{\frac{1}{2}}}$.
   
   We will use that as b_n, and the initial function as a_n. We can see that $b_n = \frac{1}{n^{\frac{1}{2}}}$ is divergent, since it's a p-series with p = 1/2.
   $\begin{align*} &\lim_{n\to \infty} \frac{a_n}{b_n}\\ &= \lim_{n\to \infty} \frac{1}{\sqrt{n}+1} \times \sqrt{n}\\ &= \lim_{n\to \infty} \frac{\sqrt{n}}{\sqrt{n}+1}\\ &= \lim_{n\to \infty} \frac{1}{1+\frac{1}{\sqrt{n}}}\\ &= 1 \end{align*}$
   
   By the definition of Comparison Test, we can determine that the series is divergent.


5. When n approaches a really large number, we can assume that the function behaves like $\frac{n^2}{n^3}=\frac{1}{n}$.
   
   We will use that as b_n, and the initial function as a_n. We can see that $b_n = \frac{1}{n}$ is divergent, since it's a p-series with p = 1.
   $\begin{align*} &\lim_{n\to \infty} \frac{a_n}{b_n}\\ &= \lim_{n\to \infty} \frac{n^2}{n^3 + \cos^2{n}} \times n\\ &= \lim_{n\to \infty} \frac{1}{1+\frac{\cos^2{n}}{n^3}}\\ &= 1 \end{align*}$
   
   By the definition of Comparison Test, we can determine that the series is divergent.

Integral Test Suppose that f is a continuous, positive and decreasing function on the interval [1,∞). If an = f(n) for n ≥ 1, then $\sum a_n$ and $\int_{1}^{\infty} f(x)\; dx$ are both divergent, or convergent.

  Example 5.5 - Integral Test

Determine whether the given sequence converges or diverges using the Integral Test.
$\begin{align*} &\text{A. } \frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \frac{1}{15} + \frac{1}{19} + ...\\ &\text{B. } \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} + \frac{1}{26} + ...\\ &\text{C. } \sum_{n=1}^{\infty} \frac{n}{(n^2+1)^{\frac{3}{2}}}\\ \end{align*}$

Solution:
Before doing the integral, we have to make sure that the series/function is positive, continuous and decreasing.

1. The general form of the series above is:
   $a_n\sum_{n=1}^{\infty} \frac{1}{4n-1}$.
   
   We can already see that this function is indeed decreasing, since the denominator increases as n increases. So, we can do the integral:
   $\begin{align*} &\int_{1}^{\infty} \frac{1}{4x-1} \; dx \\ &= \lim_{t\to \infty} \int_{1}^{t} \frac{1}{4x-1} \; dx\\ &= \lim_{t\to \infty} \left ( \frac{1}{4} \ln{|4x-1|} \right ]_{1}^{t} \\ &= \lim_{t\to \infty} \left ( \frac{\ln{|4t-1|} - \ln{|3|}}{4} \right )\\ &= \infty \end{align*}$
   
   Since we obtain infinite as the result, we can see that this series diverges.


2. The general form of the series above is:
   $a_n\sum_{n=1}^{\infty} \frac{1}{n^2+1}$.
   
   We can already see that this function is indeed decreasing, since the denominator increases as n increases. So, we can do the integral:
   $\begin{align*} &\int_{1}^{\infty} \frac{1}{x^2+1} \; dx \\ &= \lim_{t\to \infty} \int_{1}^{t} \frac{1}{x^2+1} \; dx\\ &= \lim_{t\to \infty} \left ( \tan^{-1}{x} \right ]_{1}^{t} \\ &= \lim_{t\to \infty} \left ( \tan^{-1}{t} - \tan^{-1}{1} \right )\\ &= \frac{\pi}{2} - \frac{\pi}{4}\\ &= \frac{\pi}{4} \end{align*}$
   
   Since we obtain a number that's not infinite, we can see that this series converges.


3. We can already see that this function is indeed decreasing, since the denominator overpowers the numerator. So, we can do the integral:
   $\begin{align*} &\sum_{n=1}^{\infty} \frac{n}{(n^2+1)^{\frac{3}{2}}} \\ &= \lim_{t\to \infty} \int_{1}^{t} \frac{x}{(x^2+1)^{\frac{3}{2}}} \; dx\\ \\ u&=x^2+1\\ du&=2x \; dx\\ \\ &= \lim_{t\to \infty} \int_{1}^{t} \frac{1}{2(u)^{\frac{3}{2}}} \; du\\ &= \lim_{t\to \infty} \left ( -\frac{1}{(u)^{\frac{1}{2}}} \right]_{1}^{t}\\ &= \lim_{t\to \infty} \left ( -\frac{1}{(x^2+1)^{\frac{1}{2}}} \right]_{1}^{t}\\ &= \lim_{t\to \infty} \left ( - \frac{1}{(t^2+1)^{\frac{1}{2}}} + \frac{1}{\sqrt{2}} \right )\\ &= \frac{1}{\sqrt{2}} \end{align*}$
   
   Since we obtain a number that's not infinite, we can see that this series converges.

Ratio Test Let $\sum a_n$ and $L = \lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|$. Depending on the value of L: If L < 1, the series is convergent. If L > 1, the series is divergent. If L = 1, the test is inconclusive.

  Example 5.6 - Ratio Test

Determine whether the given sequence converges or diverges using the Ratio Test.
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \frac{e^n}{n!}\\ &\text{B. } \sum_{n=1}^{\infty} \frac{n^2+1}{3^{1-2n}}\\ \end{align*}$

Solution:

1. 
   $\begin{align*} a_n&= \frac{e^n}{n!}\\ a_{n+1}&= \frac{e^{n+1}}{(n+1)!}\\ \\ L &= \lim_{n\to \infty} \left |\frac{a_{n+1}}{a_n} \right|\\ &= \lim_{n\to \infty} \left |\frac{e^{n+1} (n!)}{e^{n} (n+1)!} \right|\\ &= \lim_{n\to \infty} \left |\frac{e (n!)}{(n+1)!} \right|\\ &= \lim_{n\to \infty} \left |\frac{e}{(n+1)} \right|\\ &= 0\\ \end{align*}$
   
   By the Ratio Test, this series converges.


2. 
   $\begin{align*} a_n&= \frac{n^2+1}{3^{1-2n}}\\ a_{n+1}&= \frac{(n+1)^2+1}{3^{1-2(n+1)}}\\ \\ L &= \lim_{n\to \infty} \left |\frac{a_{n+1}}{a_n} \right|\\ &= \lim_{n\to \infty} \left |\frac{((n+1)^2+1)(3^{1-2n})}{(n^2+1)(3^{1-2(n+1)})} \right|\\ &= \lim_{n\to \infty} \left |\frac{9((n+1)^2+1)}{(n^2+1)} \right|\\ &= 9 \end{align*}$
   
   By the Ratio Test, this series diverges.

Alternating Series Test If the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1}b_n$ satisfies: $b_{n+1} \leq b_n$ $\lim_{n\to \infty} b_n = 0$ then the series converges.

  Example 5.7 - Conditional / Absolute Convergence

Determine whether the given sequence converges (conditionally or absolutely) or diverges.
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}\\ &\text{B. } \sum_{n=1}^{\infty} (-1)^{n-1} \sin{\left ( \frac{1}{n} \right )}\\ &\text{C. } \sum_{n=1}^{\infty} \frac {(-1)^n} {(\ln{n})^n}\\ \end{align*}$

Solution:

1. By the Alternating Series Test, we can already see that the series converges. However:
   $\begin{align*} &\sum_{n=1}^{\infty} \left | \frac{(-1)^{n+1}}{n+1} \right |\\ &=\sum_{n=1}^{\infty} \frac{1}{n+1}\\ \\ &\lim_{n\to \infty} \frac{1}{n+1}\\ \end{align*}$
   
   Taking the absolute value of each term of the series will make the series diverge (it's a P-series with p=1). So, this series conditionally converges.


2. By the Alternating Series Test, we can already see that the series converges. However:
   $\begin{align*} &\sum_{n=1}^{\infty} \left | (-1)^{n-1} \sin{\left ( \frac{1}{n} \right )} \right |\\ \\ &\lim_{n\to \infty} \left | (-1)^{n-1}\sin{\left ( \frac{1}{n} \right )} \right |\\ &=\lim_{n\to \infty} \sin{\left ( \frac{1}{n} \right )}\\ &=\sin{ \left ( \lim_{n\to \infty} \frac{1}{n} \right )}\\ \end{align*}$
   
   Taking the absolute value of each term of the series will make the series diverge (it's a P-series with p=1). So, this series conditionally converges.


3. The series above is indeed decreasing.
   
   By the Alternating Series Test, we can already see that the series converges. However:
   $\begin{align*} &\sum_{n=1}^{\infty} \frac {(-1)^n} {(\ln{n})^n}\\ \\ &\lim_{n\to \infty} \left | \frac {(-1)^n} {(\ln{n})^n} \right |\\ &=\lim_{n\to \infty} \frac{1}{(\ln{n})^n}\\ \end{align*}$
   
   Let's try using the Ratio Test here:
   $\begin{align*} &=\lim_{n\to \infty} \frac{(\ln{n})^{n}}{(\ln{n+1})^{n+1}}\\ &=\lim_{n\to \infty} \frac{(\ln{n})^{n}}{(\ln{(n+1)})^n(\ln{(n+1)})}\\ &=\lim_{n\to \infty} \frac{1}{\ln{(n+1)}}\\ &= 0\\ \end{align*}$
   
   So, the series is still convergent after taking its absolute value. Hence, this series is absolutely convergent.

Power Series
Power series is generally written in the form:
$\sum_{n=1}^{\infty} a_n(x-c)^n$

The power series will be convergent if $\lim_{n\to \infty} \left | \frac{u_{n+1}}{u_n} \right | < 1$ and the series will be divergent elsewhere.

  Example 5.8 - Power Series Convergence

Find the convergence interval of the given series.
$\begin{align*} &\text{A. } \sum_{n=1}^{\infty} \frac{-2(x+6)^n}{n(-5)^n}\\ &\text{B. } \sum_{n=1}^{\infty} \frac{n(x-1)^n}{(n+3)(-3)^n}\\ &\text{C. } \sum_{n=0}^{\infty} \frac{x^n}{n!}\\ \end{align*}$

Solution:

1. Let's find the interval first:
   $\begin{align*} \lim_{n\to \infty} \left | \frac{u_{n+1}}{u_n} \right | &< 1\\ \lim_{n\to \infty} \left | \frac{-2(x+6)^{n+1}}{(n+1)(-5)^{n+1}} \times \frac{n(-5)^n}{-2(x+6)^n} \right | &< 1\\ \left | \frac{(x+6)}{(-5)} \right | &< 1\\ |x+6|&<5\\ -5<x+6&<5\\ -11<x&<-1 \end{align*}$
   
   Our remaining step is to verify whether the series converge/diverge at those endpoints (-11 and -1). Let's start with -11:
   $\begin{align*} &\lim_{n\to \infty}\frac{ -2(-5)^n }{n(-5)^n}\\ &=\lim_{n\to \infty} -\frac{2}{n}\\ \end{align*}$
   
   We can already see that this series diverge (It's a P-series with p=1). Now, what about -1:
   $\begin{align*} &\lim_{n\to \infty}\frac{ -2(5)^n }{n(-5)^n}\\ &=\lim_{n\to \infty} -2\frac{(-1)^n}{n}\\ \end{align*}$
   
   By the Alternating Series Test, we can see that this series converge. Hence, our final solution is:
   $-11 < x \leq -1$


2. Let's find the interval first:
   $\begin{align*} \lim_{n\to \infty} \left | \frac{u_{n+1}}{u_n} \right | &< 1\\ \lim_{n\to \infty} \left | \frac{(n+1)(x-1)^{n+1}}{(n+4)(-3)^{n+1}} \times \frac{(n+3)(-3)^n}{n(x-1)^n} \right | &< 1\\ \left | \frac{x-1}{-3} \right | &< 1\\ |x-1|&<3\\ -3<x-1&<3\\ -2<x&<4 \end{align*}$
   
   Our remaining step is to verify whether the series converge/diverge at those endpoints (-2 and 4). Let's start with -2:
   $\begin{align*} &\lim_{n\to \infty}\frac{n(-3)^n}{(n+3)(-3)^n}\\ &=\lim_{n\to \infty} \frac{n}{n+3}\\ \end{align*}$
   
   By the Divergence Test, we can see that this series diverge (The limit of this series is 1). Now, what about 4:
   $\begin{align*} &\lim_{n\to \infty}\frac{n(3)^n}{(n+3)(-3)^n}\\ &=\lim_{n\to \infty} \frac{n (-1)^n}{n+3}\\ \end{align*}$
   
   By the Alternating Series Test, we can see that this series diverge (The limit of n/(n+3) is 1, not 0). Hence, our final solution is:
   $-2 < x < 4$


3. Let's find the interval first:
   $\begin{align*} \lim_{n\to \infty} \left | \frac{u_{n+1}}{u_n} \right | &< 1\\ \lim_{n\to \infty} \left | \frac{x^{n+1}}{(n+1)!} \times \frac{n!}{x^n} \right | &< 1\\ \left | \frac{x-1}{\infty} \right | &< 1\\ |x-1|&<\infty\\ -\infty < x-1 &< \infty\\ -\infty+1 < x& < \infty+1\\ -\infty < x &< \infty \end{align*}$
   
   This series converges for all values of x. Our final solution is:
   $-\infty < x < \infty$

Taylor's Series
y=f(x) can be written as Taylor’s series about x=a if the derivatives of all order exist and continuous at x=a:
$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$

Mac-Laurin's Series
It's Taylor's Series with a=0:
$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Some Mac-Laurin's Series
Knowing these series may come in handy.

Function	Mac-Laurin's Series
sin(x)	$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} + ...$
cos(x)	$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} + ...$
ex	$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$

  Example 5.9 - Mac-Laurin's Series I

Write the first 3 non-zero terms of the Mac-Laurin's series of the given function:
$\begin{align*} &\text{A. } f(x)=\frac{2}{3}e^{3x}\\ &\text{B. } f(x)=-x^3\cos{x}\\ &\text{C. } f(x)=\frac{1}{1-x}\\ \end{align*}$

Solution:

1. Deriving the functions:
   $\begin{align*} f(x)&=\frac{2}{3}e^{3x} \;f(0)=\frac{2}{3}\\ f'(x)&=2e^{3x} \;f'(0)=2\\ f''(x)&=6e^{3x}\;f''(0)=6\\ \end{align*}$
   
   So, we can obtain:
   $f(x) = \frac{2}{3} + 2x + 3x^2 + ...$


2. We can derive the function manually, but there's a much faster method. Recall the Mac-Laurin's series of cos(x):
   $\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} + ...\\ $
   
   What if we multiply both sides with -x³?
   $-x^3\cos{x} = -x^3 + \frac{x^5}{2!} - \frac{x^7}{4!} + ...\\ $


3. Deriving the functions:
   $\begin{align*} f(x)&=\frac{1}{1-x} \;f(0)=1\\ f'(x)&= \frac{1}{(1-x)^2} \;f'(0)=-1\\ f''(x)&= \frac{2}{(1-x)^3}\;f''(0)=2\\ \end{align*}$
   
   So, we can obtain:
   $f(x) = 1 - x + x^2 + ...$

  Example 5.10 - Mac-Laurin's Series II

Given Mac-Laurin's series, identify the function of that series:
$\begin{align*} &\text{A. } \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{n!}\\ &\text{B. } \sum_{n=0}^{\infty} -\frac{\left ( \frac{2}{3} \right )^nx^n}{n!}\\ &\text{C. } \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}\\ &\text{D. } \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)!}\\ &\text{E. } \sum_{n=0}^{\infty} 3(-1)^n \frac{x^{4n}}{(2n)!}\\ \end{align*}$

Solution:

1. Let's try re-writing the general term for the series above:
   $\begin{align*} &\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \end{align*}$
   
   Since this series looks similar to Mac-Laurin's series of e^x, we can check it out:
   $\begin{align*} e^x&=\sum_{n=0}^{\infty} \frac{x^n}{n!}\\ e^{-x} &= \sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\\ \end{align*}$
   
   So, the function with the given Mac-Laurin's series is e^-x.


2. We can see that this series is also reminiscent of Mac-Laurin's series of e^x. Now, we can manipulate that series so that we can identify the function:
   $\begin{align*} &\sum_{n=0}^{\infty} -\frac{\left ( \frac{2}{3} \right )^nx^n}{n!}\\ &= - \sum_{n=0}^{\infty} \frac{\left ( \frac{2}{3}x \right )^n}{n!}\\ &= -e^{\frac{2}{3}x} \end{align*}$
   
   So, the function with the given Mac-Laurin's series is $-e^{\frac{2}{3}x}$.


3. This series is also reminiscent of Mac-Laurin's series of e^x. Now, we can manipulate that series so that we can identify the function:
   $\begin{align*} &\sum_{n=0}^{\infty} \frac{x^{2n}}{n!}\\ &= \sum_{n=0}^{\infty} \frac{\left ( x^2 \right )^n}{n!}\\ &= e^{x^2} \end{align*}$
   
   So, the function with the given Mac-Laurin's series is e^x2.


4. From 2n+1, we can already see that this series may be related to either sin(x) or cos(x). Let's try manipulating the function:
   $\begin{align*} &\sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2n+1}}{(2n+1)!}\\ &= \sum_{n=0}^{\infty} (-1)^n(-1) \frac{x^{2n+1}}{(2n+1)!}\\ &= - \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\\ &= -\sin{x} \end{align*}$
   
   So, the function with the given Mac-Laurin's series is -sin(x).


5. From 2n, we can already see that this series may be related to either sin(x) or cos(x). Let's try manipulating the function:
   $\begin{align*} &\sum_{n=0}^{\infty} 3(-1)^n \frac{x^{4n}}{(2n)!}\\ &= 3\sum_{n=0}^{\infty} (-1)^n \frac{(x^2)^{2n}}{(2n)!}\\ &= 3\cos{x^2} \end{align*}$
   
   So, the function with the given Mac-Laurin's series is 3 cos x².