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I'm trying to figure out if there is a finite constant $C$ such that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq C$ for all $m\in\mathbb{N}$.

I can see that $\sum_{n=1}^\infty \frac{m}{(n+m)^2}\leq\sum_{n=1}^\infty \frac{m}{n^2}=mc$ for some finite constant $c$, but this is a weaker statement.

If I naively replace integers by reals and sums by integrals then I feel like the original series should be bounded.

JS1204
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3 Answers3

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Use that $\frac{m}{(n+m+1)(n+m)}= \frac{m}{(n+m)}-\frac{m}{(n+m+1)}$ for all $n,m \in \mathbb{N}$. We can then see that your series is bounded by $C=1$ by comparing it with a telescoping series. More precisley:

For every $N, m \in \mathbb{N}$ we have $$\sum_{n=1}^N \frac{m}{(n+m)^2} = \sum_{n=0}^N \frac{m}{(n+m+1)^2} \leq $$ $$ \leq \sum_{n=0}^N \frac{m}{(n+m+1)(n+m)} = \sum_{n=0}^N \frac{m}{(n+m)}-\frac{m}{(n+m+1)} = $$ $$= \frac{m}{m}-\frac{m}{(N+m+1)}=1 - \frac{m}{(N+m+1)} \leq 1 $$

Henrik
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By the integral comparison test, $\sum_{n=1}^\infty \frac{1}{(n+m)^2} =\sum_{n=m+1}^\infty \frac{1}{n^2} \approx \int_m^{\infty} \frac{dx}{x^2} =\frac{-1}{x}\big|_m^{\infty} =\frac{1}{m} $, so $\sum_{n=1}^\infty \frac{m}{(n+m)^2} \approx 1 $.

To be more precise, $\int_m^{\infty} \frac{dx}{x^2} > \sum_{n=m+1}^\infty \frac{1}{n^2} >\int_{m+1}^{\infty} \frac{dx}{x^2} $, so $\frac{1}{m} > \sum_{n=m+1}^\infty \frac{1}{n^2} >\frac{1}{m+1} $ or $1 >\sum_{n=m+1}^\infty \frac{m}{n^2} =\sum_{n=1}^\infty \frac{m}{(n+m)^2} >(1-\frac1{m+1}) $.

marty cohen
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  • Thanks. Early morning brain rot. Fixed. – marty cohen Aug 21 '15 at 13:52
  • Thanks, I needed the extra detail to understand. This is a very nice technique for dealing with summation. – JS1204 Aug 21 '15 at 14:14
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    The basic idea is that, if $f(x)$ is decreasing,$\int_n^{n+1} f(x) dx < f(n) <\int_{n-1}^{n} f(x) dx$. Then sum this.

    If $f(x)$ is increasing,$\int_{n-1}^{n} f(x) dx < f(n) <\int_{n}^{n+1} f(x) dx$.

     – marty cohen Aug 21 '15 at 14:26
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For any $x>0$: $$\sum_{n=1}^{+\infty}\frac{x}{(n+x)^2}= x\cdot \psi'(x+1) = x\cdot\frac{d^2}{dx^2}\log\Gamma(x+1)\tag{1} $$ but a way simpler bound is achieved by considering that, under the same assumption: $$ \int_{0}^{+\infty}\frac{x}{(y+x)^2}\,dy = 1.\tag{2}$$ It is interesting to point that we may use this rather trivial bound to prove some version of Stirling's inequality: it is enough to divide the RHS of $(1)$ by $x$ and integrate twice.

Jack D'Aurizio
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