# Is the following Sum inequality true?

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!

• The right side of each is $+\infty.$ Is there a typo, maybe they should be alternating series? Commented Jun 17, 2018 at 18:17
• @coffeemath better or worse now?
– Jam
Commented Jun 17, 2018 at 18:32
• 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 $\sum_{k=0}^\infty(-1)^k\dfrac{m^{2k}}{(2k+1)}!=\dfrac{\sin m}{m}$. Commented Jun 17, 2018 at 18:55
• @LutzL check again!
– Jam
Commented Jun 17, 2018 at 18:59
• 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. Commented Jun 17, 2018 at 19:03

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 $C = 1/2$.

• Perfect!!! Nice trick.
– Jam
Commented Jun 17, 2018 at 22:51
• @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 $N \geqslant 1$.
– RRL
Commented Jun 17, 2018 at 23:04