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 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 $C = 1/2$.