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Find $$\lim_{n\to\infty}\sum_{m=1}^{\infty}\frac{1}{m^2 + n^2}$$

I have the following theorems/definitions:

Theorem: (Dominated convergence theorem (DCT)). Let $(f_n)_{n\geq 1}$ be a sequence of functions such that $\lim\limits_{n\to\infty}f_n =f$ pointwise. If there exists an integrable function $g: S\to[0,\infty]$ such that $\left|f_n\right|\leq g$ for all $n\geq 1$, then $f$ is integrable and $$\lim\limits_{n\to\infty}\int_Sf_n d\mu = \int_Sf d\mu.$$

Def: Let $S = \mathbb{N}$ and $\mathcal{A} = \mathcal{P}(\mathbb{N})$. We write $\#A$ for the number of elements in a finite set $A$, and we set $\#A = \infty$ if $A$ is infinite. Let $\mu:\mathcal{A}\to[0,\infty]$ be given by $\mu(A) = \#A$. Then $\mu$ is a measure. Often $\mu$ is denoted by $\tau$ and is called the counting measure.

Exercise: Find $\lim\limits_{n\to\infty}\sum\limits_{m=1}^{\infty}\dfrac{1}{m^2 + n^2}$. Hint: use the counting measure and the DCT.

If I'm not mistaken then $\lim\limits_{n\to\infty}\dfrac{1}{m^2 + n^2} =0$ pointwise. That means that if I can think of an integrable function $g$ such that $\left|\dfrac{1}{m^2 + n^2}\right|\leq g$ for all $n\geq 1$, then $\lim\limits_{n\to\infty}\sum\limits_{m=1}^{\infty}\dfrac{1}{m^2 + n^2} = \sum\limits_{m=1}^{\infty}0\,d\mu$, by the DCT theorem. However I can't find such a function $g$ and I'm not sure how I can use the counting measure doing so..

Question: How do I find $\lim\limits_{n\to\infty}\sum\limits_{m=1}^{\infty}\dfrac{1}{m^2 + n^2}$ using the counting measure and the DCT?

Thanks in advance!

Edit: With some help from the commentsection I think I might have found the solution:

We have $\lim\limits_{n\to\infty}f_n = 0$ pointwise and $\left|f_n\right|\leq g$, $g$ integrable. Hence $\lim\limits_{n\to\infty}\int_S f_n d\mu = \int_S f d\mu$. If $\mu$ is the counting measure then $$\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty\dfrac{1}{m^2 + n^2} = \lim\limits_{n\to\infty}\int\limits_{m\in\mathbb{N}} f_n d\mu$$ Hence $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty\dfrac{1}{m^2 + n^2}= \sum\limits_{m=1}^\infty 0 \,d\mu = 0$.

Is this correct?

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    $\begingroup$ Try: $g(m) = \frac{1}{m^2}$ $\endgroup$ – Mohammad W. Alomari Dec 22 '17 at 11:42
  • $\begingroup$ You can also use formula 6.3.13 people.math.sfu.ca/~cbm/aands/page_259.htm so it is $\sim\frac{\pi}{2n}\to 0$. $\endgroup$ – zwim Dec 22 '17 at 11:54
  • $\begingroup$ Just a comment : $\sum_{m \ge 1} \frac{1}{m^2+n^2} \le \sum_{m = 1}^N \frac{1}{m^2+n^2} +\sum_{m \ge N+1} \frac{1}{m^2} $. You can choose $N$ such that $\sum_{m \ge N+1} \frac{1}{m^2} < \epsilon$, then choose $n$ such that $\sum_{m = 1}^N \frac{1}{m^2+n^2} < \epsilon$. $\endgroup$ – reuns Dec 22 '17 at 11:55
  • $\begingroup$ @mwomath I thought of that, but I'm confused as to whether or not $g$ is integrable in that case, as $\frac{1}{m^2}$ is not a continuous function if $m\in\{1,2,3,\ldots\}$ right? $\endgroup$ – titusAdam Dec 22 '17 at 12:00
  • $\begingroup$ You need Lebesgue integrable function not continuous. Follow 'reuns' hint $\endgroup$ – Mohammad W. Alomari Dec 22 '17 at 12:05
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You didn't (explicitely) check yet that the $g$, $f_n$ are measurable and that $g$ is integrable.

For the complete proof:

As mwomath stated, use $g(m)=\frac{1}{m^2}$. Both $f_n$ for all $n\geq 1$ and $g$ are measurable, because $\mathcal{A} = \mathcal{P}(\mathbb{N})$ (for any Borel-measurable subset $B$ we trivially have that $g^{-1}(B)\in \mathcal{P}(\mathbb{N})$ and $f_n^{-1}(B)\in \mathcal{P}(\mathbb{N})$).

Now we have $\int\limits_{m\in\mathbb{N}} g\ d\mu=\sum\limits_{m=1}^\infty\dfrac{1}{m^2}=\frac{\pi^2}{6}<\infty$, so $g$ is integrable and obviously $|f_n|=\frac{1}{m^2+n^2}\leq \frac{1}{m^2}=g$ for all $n\geq 1$.

Indeed $\lim\limits_{n\to\infty}f_n = 0$ pointwise, so by the DCT you get $$\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty\dfrac{1}{m^2 + n^2} = \lim\limits_{n\to\infty}\int\limits_{m\in\mathbb{N}} f_n d\mu=\int\limits_{m\in\mathbb{N}} \lim\limits_{n\to\infty}f_n d\mu=\int\limits_{m\in\mathbb{N}} 0 d\mu=0$$

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Dominating Function $$ \frac1{m^2+n^2}\le\frac1{m^2} $$ and $$ \begin{align} \sum_{m=1}^\infty\frac1{m^2} &\le1+\sum_{m=2}^\infty\frac1{m(m-1)}\\ &=2 \end{align} $$


Using Riemann Sums $$ \begin{align} \lim_{n\to\infty}\sum_{m=1}^\infty\frac1{m^2+n^2} &=\lim_{n\to\infty}\frac1n\lim_{n\to\infty}\frac1n\sum_{m=1}^\infty\frac1{\frac{m^2}{n^2}+1}\\ &=\lim_{n\to\infty}\frac1n\int_0^\infty\frac1{1+x^2}\,\mathrm{d}x\\[6pt] &=\lim_{n\to\infty}\frac\pi{2n}\\[9pt] &=0 \end{align} $$


Exact Sum and Asymptotic Approximation $$ \begin{align} \sum_{m=1}^\infty\frac1{m^2+n^2} &=-\frac1{2in}\sum_{m=1}^\infty\left(\frac1{m+in}+\frac1{-m+in}\right)\tag1\\ &=-\frac1{2n^2}-\frac1{2in}\sum_{m\in\mathbb{Z}}\frac1{m+in}\tag2\\ &=-\frac1{2n^2}-\frac1{2in}\pi\cot(\pi in)\tag3\\ &=-\frac1{2n^2}+\frac1{2n}\pi\coth(\pi n)\tag4\\ &=\frac\pi{2n}-\frac1{2n^2}+O\!\left(\frac1{ne^{2n}}\right)\tag5 \end{align} $$ Explanation:
$(1)$: partial fractions
$(2)$: rewrite the sum as a principal value sum
$(3)$: equation $(7)$ from this answer
$(4)$: $\cot(iz)=-i\coth(z)$ [Exact Sum]
$(5)$: $\coth(x)=1+O\!\left(e^{-2x}\right)$ as $x\to\infty$ [Asymptotic Approximation]

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  • $\begingroup$ @user296113: it is the product of two limits. One is the Riemann Sum approximation of $\int_0^\infty\frac{\mathrm{d}x}{1+x^2}=\frac\pi2$ and the other is $\lim\limits_{n\to\infty}\frac1n=0$. How is that not rigorous? $\endgroup$ – robjohn Dec 23 '17 at 4:53
  • $\begingroup$ Sorry, I see now it's ok and infact all this work deserve at least +1. $\endgroup$ – user296113 Dec 23 '17 at 8:37
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Because $(m+n)^2 \le 2(m^2 + n^2),$

$$\sum_{m=1}^{\infty}\frac{1}{2(m^2+n^2)} \le \sum_{m=1}^{\infty}\frac{1}{(m+n)^2} = \sum_{m=n+1}^{\infty}\frac{1}{m^2}.$$

Since $\sum_m 1/m^2<\infty,$ the last sum $\to 0$ as $n\to \infty,$ proving the limit in question is $0.$

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You can use the following identity $$ \frac{1}{m^2+n^2}=\int_{0}^{+\infty}\frac{\sin(m x)}{m}e^{-nx}\,dx $$ (which is a simple consequence of integration by parts) and recall that $$ \sum_{m\geq 1}\frac{\sin(m x)}{m} $$ is the Fourier series of a sawtooth wave, in particular a bounded function $w(x)$ (each partial sum is uniformly bounded between $-\text{Si}(\pi)$ and $\text{Si}(\pi)$). By the DCT it follows that $$ \lim_{n\to+\infty}\sum_{m\geq 1}\frac{1}{m^2+n^2}=\lim_{n\to +\infty}\underbrace{\int_{0}^{+\infty}w(x) e^{-nx}\,dx}_{O\left(\frac{1}{n}\right)}=0.$$

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