I would like to show that $$\sum_{n=1}^\infty(n+1)^p(\frac{1}{n^q}-\frac{1}{(n+1)^q})$$ converges, where $1<p<q$.

How would I go about showing that? I tested some values of $p$ and $q$ on WolframAlpha, it seems like Wolframalpha says that Ratio Test / Root Test would be inconclusive, and we need to use comparison test.

However, I am having a mental block to see what series we should compare with. I am guessing probably something like $1/n^{(q/p)}$ or even $1/n^{1+q-p}$, but I can't see it at the moment.

Thanks for any enlightenment.

  • $\begingroup$ $n^{-q}-(n+1)^{-q} = \int_n^{n+1} q x^{-q-1}dx = q n^{-q-1}+\int_n^{n+1} q (x^{-q-1}-n^{-q-1})dx$ $ = q n^{-q-1}-\int_n^{n+1} q\int_n^x (q+1)t^{-q-2}dtdx = q n^{-q-1} + \mathcal{O}(q(q+1) n^{-q-2})$ $\endgroup$ – reuns Oct 22 '16 at 21:09

Let $f(x)=x^{-q}$, so that $f^{\prime}(x)=-qx^{-q-1}$. Then by the mean value theorem, $$ \frac{1}{n^q}-\frac{1}{(n+1)^q}=f(n)-f(n+1)=-f^{\prime}(t)=qt^{-q-1}$$ for some $t\in (n,n+1)$. Therefore $$ \frac{1}{n^q}-\frac{1}{(n+1)^q}\leq qn^{-q-1}$$ hence $$ (n+1)^p\Big(\frac{1}{n^q}-\frac{1}{(n+1)^q}\Big)\leq q\frac{(n+1)^p}{n^{q+1}}\leq\frac{2^pq}{n^{1+q-p}} $$ since $n+1\leq 2n$.

Finally, the series $\sum_{n=1}^{\infty}\frac{1}{n^{1+q-p}}$ converges because $p<q$, so the original series converges by the comparison test.


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