-1

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?

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
Servaes
  • 63,261
  • 7
  • 75
  • 163
  • 3
    Your proof must be invalid: if $x=y=1$, the series diverges. – TonyK Nov 13 '16 at 15:14
  • 1
    $(k+1)^y -k^y = y\int_k^{k+1} t^{y-1} dt = y k^{y-1} +y\int_k^{k+1} (t^{y-1}-k^{y-1}) dt $ $ = y k^{y-1} +y(y-1)\int_k^{k+1} \int_k^t u^{y-2} du dt = y k^{y-1}+\mathcal{O}(y (y-1)k^{y-2})$ – reuns Nov 13 '16 at 18:55
  • Anyone care to explain the downvote? – Servaes Mar 10 '24 at 18:46

2 Answers2

3

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.

N. S.
  • 132,525
  • You don't really need limit comparison: and observe case $n=1$ as most significant to convergence.$$\begin{align}\sum_{k\ge1}\frac{(k+1)^y-k^y}{k^x}&=\sum_{k\ge1}\sum_{n\ge1}\frac{y(y-1)(y-2)\dots(y-n)}{n!}\frac{k^{y-n}}{k^x}\&=\sum_{n\ge1}\frac{y(y-1)(y-2)\dots(y-n)}{n!}\sum_{k\ge1}k^{y-n-x}\end{align}$$ – Simply Beautiful Art Nov 13 '16 at 16:30
  • @Servaes The sum over $k$ is the zeta function, which approaches $1$. So, asymptotically, the sum over $n$ converges if the sum over $k$ converges. – Simply Beautiful Art Nov 14 '16 at 14:18
0

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 ?

Nicolas FRANCOIS
  • 4,846