The first series,
[tex]\displaystyle \sum_{k=2}^\infty \frac{\cos(k)}{k^2}[/tex]
is convergent. By comparison, using the fact that |cos(x)| ≤ 1 for all real x,
[tex]\displaystyle \sum_{k=2}^\infty \frac{\cos(k)}{k^2} \le \sum_{k=2}^\infty \frac1{k^2}[/tex]
and the bounding series is a convergent p-series (with p = 2).
The second series,
[tex]\displaystyle \sum_{k=2}^\infty \frac{e^k}{\left(2+\frac1k\right)^k}[/tex]
is divergent by the limit test. We have
[tex]\dfrac{e^k}{\left(2 + \frac1k\right)^k} = \dfrac{e^k}{2^k\left(1+\frac1{2k}\right)^k} = \left(\dfrac e2\right)^k \cdot \dfrac1{\left(1+\frac1{2k}\right)^k}[/tex]
By definition,
[tex]e = \displaystyle \lim_{k\to\infty}\left(1+\frac1k\right)^k[/tex]
so that
[tex]\displaystyle \lim_{k\to\infty}\left(1+\frac1{2k}\right)^k = \lim_{k\to\infty}\sqrt{\left(1+\frac1{2k}\right)^{2k}} = \sqrt{\lim_{k'\to\infty}\left(1+\frac1{k'}\right)^{k'}} = \sqrt{e}[/tex]
so that the limit of the summand is
[tex]\displaystyle \lim_{k\to\infty} \frac{e^k}{\left(2+\frac1k\right)^k} = \frac1{\sqrt{e}} \lim_{k\to\infty} \left(\frac e2\right)^k[/tex]
but e > 2, so the limit is ∞.
