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There are some m such that $$\sum_{k=m}^{u}\frac{1}{(2k-1)^2}<\frac{2}{m^2}$$

with $\displaystyle \forall u>m$?

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    Have you tried induction or some other standard method? – 2017-01-18
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    no i don't tried – 2017-01-18
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    Could you try and show us your effort? – 2017-01-18
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    For $m=1$, the series is bounded by $\pi^2/8<2$ for all $u$. So, the answer is yes. – 2017-01-18
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    Thank's Dr. MV!!!!!!!!!!!!!!!!!!!!!!! – 2017-01-18

1 Answers 1

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This is probably off-topic.

Writing $u=m+p$ (with $p>0$), let us consider the function $$f(m,p)=\frac{2}{m^2}-\sum_{k=m}^{m+p}\frac{1}{(2k-1)^2}$$ which write $$f(m,p)=\frac{2}{m^2}-\frac{1}{4} \left(\psi ^{(1)}\left(m-\frac{1}{2}\right)-\psi ^{(1)}\left(m+p+\frac{1}{2}\right)\right)$$ where appears the polygamma function.

What it seems is that, for low values of $m$, $f(m,p)>0$ whatever $p$ could be. Below are given some limits when $p\to \infty$. $$\left( \begin{array}{ccc} m & \lim_{p\to \infty } \, f(m,p) & \approx \\ 1 & 2-\frac{\pi ^2}{8} & +0.766299 \\ 2 & \frac{3}{2}-\frac{\pi ^2}{8} & +0.266299 \\ 3 & \frac{4}{3}-\frac{\pi ^2}{8} & +0.099633 \\ 4 & \frac{2297}{1800}-\frac{\pi ^2}{8} & +0.042411 \\ 5 & \frac{13798}{11025}-\frac{\pi ^2}{8} & +0.017819 \\ 6 & \frac{245963}{198450}-\frac{\pi ^2}{8} & +0.005720 \\ 7 & \frac{14803024}{12006225}-\frac{\pi ^2}{8} & -0.000755 \\ 8 & \frac{79817814617}{64929664800}-\frac{\pi ^2}{8} & -0.004404 \\ 9 & \frac{166001126}{135270135}-\frac{\pi ^2}{8} & -0.006518 \\ 10 & \frac{1437785878979}{1172792070450}-\frac{\pi ^2}{8} & -0.007749 \end{array} \right)$$

So, for $m>6$, it exists a value of $p$ which makes $f(m,p) <0$.