Ι got a feeling that $$\sum_{m=1}^{N}\Big\lvert \sum_{k=0}^{\infty} \frac{m^{2k}}{(2k+1)!}(-1)^{k}\Big\rvert \geq C \sum_{m=1}^{N} \frac{1}{m} $$ Does it exist a $n_o$ such that for every $N\geq n_0$ the above is true?

$$\sum_{m=1}^{N} \Big\lvert 1-\frac{m^2}{3!}+\frac{m^4}{5!}... \Big\rvert \geq C+\frac{C}{2}+\frac{C}{3}... + \frac{C}{N}$$ i feel that somehow terms will get canceled for big enough N, but i cant prove it!! ( $ m \in N$) and $0< C<1$ constant.

this came up as a part of problem i was solving . I got no idea if the above inequallity is true got no clue how to approach it!

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    $\begingroup$ The right side of each is $+\infty.$ Is there a typo, maybe they should be alternating series? $\endgroup$
    – coffeemath
    Commented Jun 17, 2018 at 18:17
  • $\begingroup$ @coffeemath better or worse now? $\endgroup$
    – Jam
    Commented Jun 17, 2018 at 18:32
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    $\begingroup$ The second formula is not the same as the first one. It should be $1-\dfrac{m^2}2+\dfrac{m^4}{3!}-\dfrac{m^6}{4!}+...$ inside the absolute value. Or the sum in the first formula should be $\sum_{k=0}^\infty(-1)^k\dfrac{m^{2k}}{(2k+1)}!=\dfrac{\sin m}{m}$. $\endgroup$ Commented Jun 17, 2018 at 18:55
  • $\begingroup$ @LutzL check again! $\endgroup$
    – Jam
    Commented Jun 17, 2018 at 18:59
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    $\begingroup$ Yes, now the formulas are the same. So you want to claim $$\sum_{m=1}^N\frac{|\sin m|}m\ge C\sum_{m=1}^N\frac1m.$$ This could be complicated as $|\sin m|$ can be arbitrarily small, the sequence of the $|\sin m|$ is dense in $[0,1]$. Thus you need some kind of avaraging argument where the larger values balance out the small values. $\endgroup$ Commented Jun 17, 2018 at 19:03

1 Answer 1


The inner sum is $\sin m /m$ and

$$\sum_{m=1}^N \frac{|\sin m|}{m} \geqslant \sum_{m=1}^N \frac{|\sin m|^2}{m} = \frac{1}{2}\sum_{m=1}^N \frac{1}{m} - \sum_{m=1}^N \frac{\cos 2m}{2m}$$.

Since the second series on the RHS converges (by the Dirichlet test) we have

$$\sum_{m=1}^N \frac{|\sin m|}{m} \geqslant \frac{1}{2}\sum_{m=1}^N \frac{1}{m} - K,$$

where $K \approx -0.2603$ for sufficiently large $N$ and your result holds for $C = 1/2$.

  • $\begingroup$ Perfect!!! Nice trick. $\endgroup$
    – Jam
    Commented Jun 17, 2018 at 22:51
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    $\begingroup$ @ManolisLyviakis: Your welcome. I made a minor edit since the LHS is eventually greater than what is shown with $K \approx -0.2603$ (the sum to four decimal places) but not for all $N \geqslant 1$. $\endgroup$
    – RRL
    Commented Jun 17, 2018 at 23:04

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