# Math Help - Evaluating Series.

1. ## Evaluating Series.

Find the sum of each of the following series,

(i) $\sum^\infty_{n=1} nx^n$
(ii) $\sum^\infty_{n=1} \frac{n}{2^n}$
(iii) $\sum^\infty_{n=1} n(n-1)x^n$

Use Taylor series to evaluate the following limits:
(a) $\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}x^2 - cos x}{x^4}$
(b) $\lim_{x \rightarrow 0} \frac{x - arctan x}{x^3}$

2. Originally Posted by panda*
Find the sum of each of the following series,

(i) $\sum^\infty_{n=1} nx^n$
(ii) $\sum^\infty_{n=1} \frac{n}{2^n}$
(iii) $\sum^\infty_{n=1} n(n-1)x^n$

Use Taylor series to evaluate the following limits:
(a) $\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}x^2 - cos x}{x^4}$
(b) $\lim_{x \rightarrow 0} \frac{x - arctan x}{x^3}$
1. Consider $\sum_{n=0}^{\infty}x^n$. Because its interval of convergence is $(-1,1)$ on that same interval it is uniformly convergent. Also note that $\forall{x}\in(-1,1)~\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$. So now since we know this series is uniformly convergent we may say this \begin{aligned}\left(\frac{1}{1-x}\right)'&=\frac{d}{dx}\sum_{n=0}^{\infty}x^n\\
&=\sum_{n=1}^{\infty}nx^{n-1}\\
&\implies{x}\cdot\left(\frac{1}{1-x}\right)'=\sum_{n=0}^{\infty}nx^n~\forall{x}\in(-1,1)
\end{aligned}

2. Direct application of number one

3. Using the same agrument as before we can see that

$x^2\cdot\left(\frac{1}{1-x}\right)''=\sum_{n=0}^{\infty}n(n-1)x^n$

3. Using the everywhere convergent Maclaurin series we may rewrite our limit as follows

\begin{aligned}\lim_{x\to{0}}\frac{1-\frac{x^2}{2}-\cos(x)}{x^4}&=\frac{1-\frac{x^2}{2}-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\right)}{x^4}\\
&=\lim_{x\to{0}}\frac{\frac{-x^4}{24}+\cdots}{x^4}\\
&=\frac{-1}{24}
\end{aligned}

4. We know that $\forall{x}\in(-1,1]~\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$. So since if $S=\left\{x:x\to{0}\right\}$ we can see that $S\subset(-1,1]$. Therefore we may rewrite our limit as

\begin{aligned}\lim_{x\to{0}}\frac{x-\arctan(x)}{x^3}&=\lim_{x\to{0}}\frac{x-\left(x-\frac{x^3}{3}+\cdots\right)}{x^3}\\
&=\lim_{x\to{0}}\frac{\frac{x^3}{3}-\cdots}{x^3}\\
&=\frac{1}{3}
\end{aligned}

3. Originally Posted by Mathstud28
[snip]

3. Using the everywhere convergent Maclaurin series we may rewrite our limit as follows

\begin{aligned}\lim_{x\to{0}}\frac{1-\frac{x^2}{2}-\cos(x)}{x^4}&=\frac{1-\frac{x^2}{2}-\left(1-\frac{x^2}{{\color{red}2}}+\frac{x^4}{24}-\cdots\right)}{x^4}\\
&=\lim_{x\to{0}}\frac{\frac{-x^4}{24}+\cdots}{x^4}\\
&=\frac{-1}{24}
\end{aligned}

[snip]
Not trying to be super picky, but shouldn't that be a two? Otherwise, the limit would diverge...

--Chris

4. Originally Posted by Chris L T521
Not trying to be super picky, but shouldn't that be a two? Otherwise, the limit would diverge...

--Chris
Yes, it was a typo. Thank you.