Test For Convergence

Question

$$ \sum_{k = 1}^\infty \left(\frac{\mathrm{e}^{2\sqrt{k}}}{\mathrm{e}^\sqrt{k}+9}\right)^{2k} $$

Answer 1

In its current form, you do not even need to use tests. Note that for $k\to\infty$, we have that $e^{2\sqrt{k}}/\left(e^{\sqrt{k}}+9\right)\to\infty$, which even more so means that

$$\lim_{k\to\infty}\left(\frac{e^{2\sqrt{k}}}{e^{\sqrt{k}}+9}\right)^{2k}=\infty$$

Since the general term of your series does not tend to $0$, your series does not fulfill the necessary conditions for convergence. Hence, it diverges.

Answer 2

Edit: The first part deals with the sequence as it once appeared to be. The remark at the end deals with the sequence as it is at this moment. For that, there is obvious divergence. One should always first ask oneself how a sequence behaves, not what test to use.

Old: We make an assumption about the $a_k$ for which we are studying $\sum a_k$. At one stage this was the series.

Let us suppose that what is intended is $$a_k=\frac{e^{2\sqrt{k}}}{(e^{\sqrt{k}}+9)^{2k} }.$$ Then $a_k$ is less than the term we get by dropping the $9$, so $$a_k\lt \frac{1}{e^{\sqrt{k}(2k-2)}}.$$ The bottom is, for $k\gt 1$, bigger than $e^{k\sqrt{k}}$ But $$\sum \frac{1}{e^{k\sqrt{k}}}$$ converges, by comparison with the geometric series $\sum \frac{1}{e^k}$. Lots of slack.

Remark: If instead $$a_k =\left(\frac{e^{2\sqrt{k}}}{e^{\sqrt{k}}+9 } \right)^{2k},$$ as it currently looks like, then note that almost immediately we have $a_k\gt 1$, so, as pointed out in a comment, there is automatic divergence.

By the way, for this version of the series the Root Test works easily. But it is not the best way to approach the problem.