[sp]Start with the Taylor series expansion $\ln(1-t) = -t - \dfrac{t^2}2 - \dfrac{t^3}3 - \ldots$ (valid when $|t|<1$).greg1313 said:Evaluate $$\sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$
my solution:greg1313 said:Evaluate $$\sum_{n=1}^\infty\frac{1}{2nx^{2n}}$$
lfdahl said:My solution:
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.
Albert said:my solution:
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:
$$\dfrac {1}{2x^2-1}<S=\sum_{n=1}^\infty\dfrac{1}{2nx^{2n}}<\dfrac {1}{2x^2-2}$$
Albert said:my solution:
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}$$
greg1313 said:Hi Albert. I don't see how the above provides a solution. Can you clarify?
The purpose of evaluating this infinite series is to determine its convergence or divergence. This information is important in various areas of mathematics and physics.
The general term of this series is given by $\frac{1}{2nx^{2n}}$, where n is the index of summation.
The series converges if the absolute value of the general term, $\frac{1}{2nx^{2n}}$, approaches 0 as n approaches infinity. This can be shown using the ratio test for convergence.
The series converges for all x values such that $|x| > 1$. This can be shown by taking the limit of the general term as n approaches infinity and using the fact that $|x| > 1$ implies $|\frac{1}{x}| < 1$.
This infinite series appears in various mathematical and physical problems, such as in the study of complex numbers, electrical engineering, and quantum mechanics. It also has connections to the Riemann zeta function and the distribution of primes.