Is there a cleaner proof of convergence for this almost-telescoping series

Question

I was wondering how the convergence of the series $$\sum_{k\geq1}\frac{(k+1)^y-k^y}{k^x},$$ depends on $x$ and $y$, where $x$ and $y$ are real numbers. I've been able to show that it converges if and only if $y\leq x$, by distinguishing nine cases; whether $x$ and $x-y$ are positive, negative or zero. If desired I can include a sketch of my proof. My question is whether there is a proof that works for all real $x$ and $y$? Or at least a proof that seems less artificial?

Answer 1

If $y\ne0$, $$(k+1)^y-k^y=k^y\left[(1+\frac1k)^y-1\right]\sim yk^{y-1}$$ so the term of the sum is equivalent to $yk^{y-x-1}$, just apply Riemann criterion (and separate special cases if any) : the series converges if and only if $y-x-1<-1$, so $x>y$.

When $y=0$, $\frac{(k+1)^y-k^y}{k^x}=0$, so the series is convergent whater $x$ is.

Is that OK for you, @TonyK ?

Answer 2

Use the Limit Comparison test with $$\sum_{k=1}^\infty \frac{1}{k^{x-y+1}}$$

$$\lim_k \frac{\frac{(k+1)^y-k^y}{k^x}}{\frac{1}{k^{x-y+1}}}=\lim_k \frac{(k+1)^y-k^y}{k^x}\frac{k^{x-y+1}}{1}=\lim_k \frac{(k+1)^y-k^y}{k^{y-1}}\\ =\lim_k \frac{(1+\frac{1}{k})^y-1}{\frac{1}{k}}$$

Now, $$\lim_{x\to 0} \frac{(1+x)^y-1}{x}= \frac{d}{dx}(1+x)^y|_{x=0}=y$$

Therefore $$\lim_k \frac{(1+\frac{1}{k})^y-1}{\frac{1}{k}}=y$$

Conclusion

If $y \neq 0$ , by the limit comparison test, the series is convergent if $x-y-1>1$ and divergent if $x-y-1 \leq 1$.

If $y=0$ your series is $\sum_{k\geq 0} 0$ is always convergent.