Does this sequence converge? Alternating and exponential

Question

$$\sum_{k=1}^{\infty}\left(-1\right)^k\frac{\left(k+1\right)^{k+1}}{k^{k+2}}$$

I started to use Dirichlet's test. However, the latter half does not decrease to 0. I am unsure of what to do.

Answer 1

Hint

Consider $$a_k=\frac{(k+1)^{k+1}}{k^{k+2}}\implies \log(a_k)=(k+1)\log(k+1)-(k+2)\log(k)$$ Now, make $\log(a_{k+1})-\log({a_k})$ and use Taylor expansions for large values of $k$. This should give $$\log(a_{k+1})-\log({a_k})=-\frac{1}{k}+O\left(\frac{1}{k^2}\right)$$ $$\frac{a_{k+1} } {{a_k} }= e^{\log(a_{k+1})-\log({a_k})}=1-\frac{1}{k}+O\left(\frac{1}{k^2}\right)$$ and you face an alternating series.

Answer 2

You can see that it is an alternating series $\sum_{k=1}^{\infty}(-1)^k a_k$ with $a_k \searrow 0$ using the following fact:

• $(1+\frac{1}{k})^{k+1}$ is decreasing (proof) with its know limit $e$.

Hence, you get

$$a_k = \frac{\left(k+1\right)^{k+1}}{k^{k+2}} =\frac{1}{k}\left(\frac{k+1}{k}\right)^{k+1}= \frac{1}{k}\underbrace{\left(1+\frac{1}{k}\right)^{k+1}}_{\searrow\; e} \Rightarrow a_k \searrow \; 0$$

So, you conclude the convergence of your series using the alternating series test.