I'm prette sure the first sum diverges. One can check that
$\displaystyle e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for integer $\displaystyle k≥3$.
Now, the harmonic series
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges, and so
$\displaystyle \sum_{n=3}^{\infty}\frac{1}{n}$ diverges as well.
but since $\displaystyle e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for $\displaystyle k≥3$
then
$\displaystyle \sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]\gt\sum_{n=3}^{\infty}\frac{1}{k}$
and so by the comparison test, both sums diverge, and since
$\displaystyle \sum_{n=1}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]=(e-2)+(e-\frac{9}{4})+\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]$,
your series in problem 1 diverges.
--Kevin C.