For what values of $\alpha$ does the series $\sum \frac{k^{-\alpha}}{1+\alpha^{-k}}$ converge? I don't think the series:
$$\sum \frac{k^{-\alpha}}{1+\alpha^{-k}}$$
converges for $\alpha \leq 0$ since the terms don't even converge to zero for those values of $\alpha$.
I have tried to do a ratio test where I get that the series converges if:
$$0 < \lim_{k \to \infty}\frac{1+\alpha^{-k}}{1+\alpha^{-(k+1)}}<1$$
But how can I find the values of $\alpha$ for which this is true? It appears that if $0 < \alpha < 1$ then the limit is $\alpha$ but if $\alpha \geq 1$ then it is $1$.
How can this be proven?
 A: Let $u_k = \frac{1}{1 + \alpha^{-k}}$ and $v_k = k^{-\alpha}$. Consider the behaviour of $(u_k)$ and $(v_k)$ for different values of $\alpha$ separately.
For $\lvert\alpha\rvert > 1$, we have $\alpha^{-k}\to 0$ and hence $u_k \to 1$. Therefore the series $\sum_k u_k\cdot v_k$ converges if and only if $\sum_k v_k$ converges for these $\alpha$. It is well-known that the series $\sum_k k^{-\alpha}$ is convergent if and only if $\alpha > 1$, so the series $\sum_k \frac{k^{-\alpha}}{1 + \alpha^{-k}}$ converges for $\alpha > 1$ and diverges for $\alpha < -1$.
For $\alpha = 1$, we have $u_k = \frac{1}{2}$ for all $k$, so the series is half the harmonic series, which diverges. For $\alpha = -1$, we have $1 + \alpha^{-k} = 0$ for odd $k$, so the terms of the series aren't even all defined then, thus we cannot properly speak of the convergence or divergence of the series $\sum_k \frac{k^{-\alpha}}{1+\alpha^{-k}}$ for $\alpha = -1$.
For $\alpha = 0$, the terms of the series as written aren't defined as well, but if we rewrite $u_k$ as $\frac{\alpha^k}{1 + \alpha^k}$, or by abuse of notation interpret $\alpha^{-k}$ as $\infty$ and then $\frac{k^{-\alpha}}{1 + \infty}$ as $0$, we get a series all of whose terms are $0$, which is clearly convergent.
For $0 < \lvert\alpha\rvert < 1$, we have $u_k \sim \alpha^k$, i.e. $u_k$ converges to $0$ exponentially, and $v_k$ grows at most polynomially, so $u_k\cdot v_k$ also converges to $0$ exponentially and the series converges. In this case, we can use the ratio or root tests to quantify the informal summary:
$$\biggl\lvert\frac{u_{k+1}v_{k+1}}{u_kv_k}\biggr\rvert = \biggl\lvert\frac{(k+1)^{-\alpha}\alpha^{k+1}}{1+\alpha^{k+1}}\cdot \frac{1 + \alpha^k}{k^{-\alpha}\alpha^k}\biggr\rvert = \lvert\alpha\rvert\cdot \biggl(\frac{k}{k+1}\biggr)^{\alpha}\cdot \frac{1+\alpha^k}{1+\alpha^{k+1}} \to \lvert\alpha\rvert,$$
so by the ratio test the series is (absolutely) convergent for $0 < \lvert\alpha\rvert < 1$.
Summing up, the series converges for $\alpha > 1$ and for $0 < \lvert\alpha\rvert < 1$, in the "right" interpretation also for $\alpha = 0$. The series diverges for $\alpha = 1$ and for $\alpha < -1$, it is not well-defined for $\alpha = -1$ (and for $\alpha = 0$ in the literal interpretation).
