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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?

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    $\begingroup$ 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... $\endgroup$ – Clement C. Oct 12 '17 at 1:52
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Using the monotone convergence theorem, we can switch the order of the limit and the sum.

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

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