# Infinite sum of expression as exponent goes to infinity

I was messing around with this sum

$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^{2n}}$

And started increasing exponent and getting numerical solutions for n=10...100....26572... you get the idea. So I noticed as n gets bigger, the sum seems to approach 1, i.e:

$\lim_{n \to \infty} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{2n}} =^{?} 1$

Is this the case? Does the limit really converge to 1?

• The first term is $1$ for all $n$; all the other ones go to zero. Now, find a theorem that allows you to swap limit and summation... – Clement C. Oct 12 '17 at 1:52

\begin{align} \lim_{n\to\infty}\sum_{k=1}^\infty \frac1{(2k-1)^{2n}}&= \lim_{n\to\infty}\left(1+\sum_{k=2}^\infty \frac1{(2k-1)^{2n}}\right)\\ &= 1+\sum_{k=2}^\infty \lim_{n\to\infty} \frac1{(2k-1)^{2n}}\\ &= 1+\sum_{k=2}^\infty 0=1\end{align}