Evaluating $\sum_{n=1}^\infty \frac{1}{2nx^{2n}}$

[Greg]

Evaluate $$\sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$

 

[Opalg]

Evaluate $$\sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$

[sp]Start with the Taylor series expansion $\ln(1-t) = -t - \dfrac{t^2}2 - \dfrac{t^3}3 - \ldots$ (valid when $|t|<1$).

Substitute $t = \dfrac1{x^2}$: $\ln\Bigl(1-\dfrac1{x^2}\Bigr) = -\dfrac1{x^2} - \dfrac1{2x^4} - \dfrac1{3x^6} - \ldots.$

Therefore $$-\dfrac12\ln\Bigl(1-\dfrac1{x^2}\Bigr) = \dfrac1{2x^2} + \dfrac1{4x^4} + \dfrac1{6x^6} + \ldots = \sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$ (the sum converging when $|x|>1$).

[/sp]

 

[Albert1]

Evaluate $$\sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$

my solution:

Spoiler

let:
$S=(\dfrac{1}{2x^2}+\dfrac{1}{4x^4}+\dfrac{1}{6x^6}+-----+\dfrac{1}{2nx^{2n}}+---)\\
P=\dfrac{1}{2}(\dfrac{1}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^6}+-----+\dfrac{1}{x^{2n}}+---)\\
Q=(\dfrac{1}{2^1x^2}+\dfrac{1}{2^2x^4}+\dfrac{1}{2^3x^6}+-----+\dfrac{1}{2^nx^{2n}}+---)$
$by \,\,"Alembert's \,\, ratio \,\,test" :$
$$\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n}=\dfrac {1}{x^2}<1, or ,x^2>1$$
$\rightarrow \left | x \right |>1$ then $S$ converges
and $\left | x \right |<1,$ $S$ diverges
if $\left | x \right |=1$ then $S=\dfrac{1}{2}(\dfrac{1}{1}+\dfrac{1}{2}+--\dfrac{1}{n}+--)$ diverges
I will focus on $\left | x \right |>1,$and find the range of $S$
for:$Q<S<P$
By infinite geometric series it is easy to get:
$$\dfrac {1}{2x^2-1}<S=\sum_{n=1}^\infty\dfrac{1}{2nx^{2n}}<\dfrac {1}{2x^2-2}$$

 

[lfdahl]

My solution:

Spoiler

Finding the limit via the derivative of the sum:

\[\frac{\mathrm{d} }{\mathrm{d} x}\sum_{n=1}^{\infty}\frac{1}{2n}x^{-2n}=-\frac{1}{x}\sum_{n=1}^{\infty}(x^{-2})^n = -\frac{1}{x}\left ( \sum_{n=0}^{\infty}(x^{-2})^n-1 \right ) \\\\ =^*-\frac{1}{x}\left (\frac{1}{1-x^{-2}}-1 \right )=-\frac{1}{x}\cdot \frac{1}{x^2-1} \\\\ \Rightarrow \sum_{n=1}^{\infty}\frac{1}{2n}x^{-2n} = -\int \frac{1}{x}\cdot \frac{1}{x^2-1}dx=^{**}-\frac{1}{2}\ln (1-x^{-2})\](*). The geometric sum converges $iff$ $x^{-2} < 1$ or $ |x| > 1$.
(**). Solving the integral:

\[-\int \frac{1}{x}\cdot \frac{1}{x^2-1}dx=-\int \frac{1}{x^3-x}dx = -\int \frac{x^{-3}}{1-x^{-2}}dx \]

Substitution with \[u = 1-x^{-2} \rightarrow -\frac{1}{2}du = -x^{-3}dx\], gives the result.

 

[Albert1]

My solution:

Spoiler

Finding the limit via the derivative of the sum:

\[\frac{\mathrm{d} }{\mathrm{d} x}\sum_{n=1}^{\infty}\frac{1}{2n}x^{-2n}=-\frac{1}{x}\sum_{n=1}^{\infty}(x^{-2})^n = -\frac{1}{x}\left ( \sum_{n=0}^{\infty}(x^{-2})^n-1 \right ) \\\\ =^*-\frac{1}{x}\left (\frac{1}{1-x^{-2}}-1 \right )=-\frac{1}{x}\cdot \frac{1}{x^2-1} \\\\ \Rightarrow \sum_{n=1}^{\infty}\frac{1}{2n}x^{-2n} = -\int \frac{1}{x}\cdot \frac{1}{x^2-1}dx=^{**}-\frac{1}{2}\ln (1-x^{-2})\](*). The geometric sum converges $iff$ $x^{-2} < 1$ or $ |x| > 1$.
(**). Solving the integral:

\[-\int \frac{1}{x}\cdot \frac{1}{x^2-1}dx=-\int \frac{1}{x^3-x}dx = -\int \frac{x^{-3}}{1-x^{-2}}dx \]

Substitution with \[u = 1-x^{-2} \rightarrow -\frac{1}{2}du = -x^{-3}dx\], gives the result.

fabulous!

 

[Greg]

my solution:

Spoiler

let:
$S=(\dfrac{1}{2x^2}+\dfrac{1}{4x^4}+\dfrac{1}{6x^6}+-----+\dfrac{1}{2nx^{2n}}+---)\\
P=\dfrac{1}{2}(\dfrac{1}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^6}+-----+\dfrac{1}{x^{2n}}+---)\\
Q=(\dfrac{1}{2^1x^2}+\dfrac{1}{2^2x^4}+\dfrac{1}{2^3x^6}+-----+\dfrac{1}{2^nx^{2n}}+---)$
$by \,\,"Alembert's \,\, ratio \,\,test" :$
$$\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n}=\dfrac {1}{x^2}<1, or ,x^2>1$$
$\rightarrow \left | x \right |>1$ then $S$ converges
and $\left | x \right |<1,$ $S$ diverges
if $\left | x \right |=1$ then $S=\dfrac{1}{2}(\dfrac{1}{1}+\dfrac{1}{2}+--\dfrac{1}{n}+--)$ diverges
I will focus on $\left | x \right |>1,$and find the range of $S$
for:$Q<S<P$
By infinite geometric series it is easy to get:
$$\dfrac {1}{2x^2-1}<S=\sum_{n=1}^\infty\dfrac{1}{2nx^{2n}}<\dfrac {1}{2x^2-2}$$

Hi Albert. I don't see how the above provides a solution. Can you clarify?

 

[Albert1]

my solution:

Spoiler

let:
$S=(\dfrac{1}{2x^2}+\dfrac{1}{4x^4}+\dfrac{1}{6x^6}+-----+\dfrac{1}{2nx^{2n}}+---)$$by \,\,"Alembert's \,\, ratio \,\,test" :$
$$\lim_{n\rightarrow \infty} \dfrac{a_{n+1}}{a_n}=\dfrac {1}{x^2}<1, or ,x^2>1$$
$\rightarrow \left | x \right |>1$ then $S$ converges
and $\left | x \right |<1,$ $S$ diverges
if $\left | x \right |=1$ then $S=\dfrac{1}{2}(\dfrac{1}{1}+\dfrac{1}{2}+--\dfrac{1}{n}+--)$ diverges
I will focus on $\left | x \right |>1,$and find the range of $S$
for:$Q<S<P$
By infinite geometric series it is easy to get:
$$\dfrac {1}{2x^2-1}<S=\sum_{n=1}^\infty\dfrac{1}{2nx^{2n}}<\dfrac {1}{2x^2-2}$$

Hi Albert. I don't see how the above provides a solution. Can you clarify?

Spoiler

$S=(\dfrac{1}{2x^2}+\dfrac{1}{4x^4}+\dfrac{1}{6x^6}+-----+\dfrac{1}{2nx^{2n}}+---)$
$P=\dfrac{1}{2}(\dfrac{1}{x^2}+\dfrac{1}{x^4}+\dfrac{1}{x^6}+-----+\dfrac{1}{x^{2n}}+---)$
$Q=(\dfrac{1}{2^1x^2}+\dfrac{1}{2^2x^4}+\dfrac{1}{2^3x^6}+-----+\dfrac{1}{2^nx^{2n}}+---)$
$$\dfrac {1}{2x^2-1}<S=-\dfrac{1}{2}ln(1-\dfrac{1}{x^2})=\sum_{n=1}^\infty\dfrac{1}{2nx^{2n}}<\dfrac {1}{2x^2-2}$$
$( \left | x \right |>1)$
(from Opalg's solution):$S=-\dfrac{1}{2}ln(1-\dfrac{1}{x^2})$
$P,Q$ are two infinite geometric series with ratio $\dfrac {1}{x^2} $ and $\dfrac{1}{2x^2}$ repectively
Opalg already has given the answer , I only give the range of $S$ ,infact they are quite close

 

[Greg]

Thanks for your participation. :)

My solution is effectively the same as lfdahl's, so I won't post it here.