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I want to compute $$\lim_{x\to \infty }\sum_{n=1}^\infty \frac{1}{n(n+x)}.$$

Can I do as follow? Consider the measurable space $(\mathbb N,\mathcal P(\mathbb N),\mu)$ where $\mu(A)=\#A$. Then, $$\sum_{n=1}^\infty \frac{1}{n(n+x)}=\int_{\mathbb N}\frac{1}{n(n+x)}d\mu(n).$$ Suppose $|x|\geq 1$. Then $$\left|\frac{1}{n(n+x)}\right|\leq \frac{1}{n(n+1)}\in L^1(\mathbb N),$$ and thus, using DCT, we finally obtain $$\lim_{x\to \infty }\sum_{n=1}^\infty \frac{1}{n(n+x)}=\sum_{n=1}^\infty \lim_{x\to \infty }\frac{1}{n(n+x)}=0.$$

Does it work ?

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  • $\begingroup$ Don’t know if your method works, but I would use M test to prove uniform convergence and then take limit termwise. $\endgroup$ – Szeto Jul 22 '18 at 13:19
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    $\begingroup$ The dominated convergence theorem is for sequences of functions. In your case you have a limit over the real numbers. You can apply your argument to deduce that for every monotonic sequence $x_m\to\infty$ you have $\lim_{m\to\infty}\sum_{n=1}^{\infty}\frac{1}{n(n+x_m)}=0$. Then you need to deduce from that that $\lim_{x\to\infty}\sum_{n=1}^{\infty}\frac{1}{n(n+x)}=0$. $\endgroup$ – user574889 Jul 22 '18 at 13:23
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    $\begingroup$ It is important to remember that the dominated convergence theorem is for sequences because of this $\endgroup$ – user574889 Jul 22 '18 at 13:27
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    $\begingroup$ This is a good approach. Use it to show that the limit when $x\to\infty$, $x$ integer, is zero, then extend the result to $x\to\infty$, $x$ real, by monotonicity of the sum with respect to $x$. $\endgroup$ – Did Jul 22 '18 at 13:30
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    $\begingroup$ @user380364 Monotonicity is not needed. What is important is that on $\mathbb{R}$ a limit $\lim_{x\to a}$ exists (as a unique limit) if and only if it exists for all sequences $x_n\to a$. This is ultimately due to $\mathbb{R}$ being first countable. $\endgroup$ – user574889 Jul 22 '18 at 14:13
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Yes, your proof is fine (at least modulo cactus's comment: You really do need to reduce to a sequence before you can apply DCT. This is no big deal, people apply DCT to $\lim_{x\to\infty}$ all the time, because it's clear that $\lim_{x\to \infty}I_x=I$ if and only if $\lim_{n\to\infty}I_{x_n}=I$ for every sequence $x_n$ with $x_n\to\infty$.)

Of course there's a very simple more elementary argument here, which I write out because it leads to Something Interesting at the bottom:

Elementary Argument: Let $\epsilon>0$. Choose $N$ so $$\sum_{n=N+1}^\infty\frac1{n^2}<\epsilon.$$Now for every $x>0$ we have $$\sum_1^\infty\frac1{n(n+x)}<\epsilon+\sum_1^N\frac1{n(n+x)};$$since $N$ is fixed it follows that $\sum_1^\infty<2\epsilon$ if $x$ is large enough.

Something Interesting: Exercise 1 Generalize the argument above to give an elementary proof of DCT for coutning measure on $\Bbb N$.

Exercise 2 Show that DCT for a general measure space follows from Egoroff's Theorem, by an argument analogous to the argument above.

I like the proof of DCT via Egoroff - at least to me it gives a much better picture of "why it's really true" then the proof from Fatou's Lemmma that you see in all the books.

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  • $\begingroup$ But why we always pass by a sequence to apply DCT ? Let's call "Corollary of DCT" the fact that $\lim_{x\to \infty }\int_{\mathbb R}f(t,x)dt=\int_{\mathbb R}\lim_{x\to\infty } f(x,t)dt$ always work ? (I made a proof in the comment of my question). I don't see in what we must pass with a sequence. Yes DCT with Egoroff's theorem is more intuitive since in fact all sequence $(f_n)$ that converge p.p. to $f$ converge almost uniformly. $\endgroup$ – user380364 Jul 22 '18 at 13:51
  • $\begingroup$ Officially we need to pass to sequences because DCT applies to sequences. In fact people do apply your "Corollary to DCT" all the time without mentioning it, because it's more or less obvious how to reduce to sequences in that case. But as cactus points out, there are other notions of limit that cannot be reduced to sequences, and for which in fact DCT fails. $\endgroup$ – David C. Ullrich Jul 22 '18 at 14:02
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As pointed out in comments, the OP's approach is largely correct; it's only flaw is that the DCT is usually stated for sequences of functions, not a continuum of them, so the proof needs an extra step along the lines of

$$\lim_{x\to\infty}\sum{1\over n(x+x)}\le\lim_{x\to\infty}\sum{1\over n(x+\lfloor x\rfloor)}=\lim_{m\to\infty}\sum{1\over n(n+m)}$$

(where $m$ is understood as an integer variable).

So as an exercise in understanding the DCT, this is a nice example.

For what it's worth, here is another alternative approach to showing the limit is $0$: By the Arithmetic-Geometric Mean inequality, we have $n+x\ge2\sqrt{nx}$, and thus

$$\sum{1\over n(n+x)}\le{1\over2\sqrt x}\sum{1\over n^{3/2}}$$

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  • $\begingroup$ (+1) This is clear and concise, and embeds an elementary way forward using the AM-GM. Well done! $\endgroup$ – Mark Viola Jul 22 '18 at 15:49
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In the same spirit as A. Pongrácz in his/her answer, after partial fraction decomposition $$S_x=\sum_{n=1}^\infty \frac{1}{n(n+x)}=\frac{H_x}{x}$$ Using the asymptotics of harmonic numbers, we then have $$S_x=\frac{\gamma +\log \left({x}\right)}{x}+\frac{1}{2 x^2}-\frac{1}{12 x^3}+O\left(\frac{1}{x^5}\right)$$ which seems to be quite good even for small values of $x$.

Forexample $$S_{10}=\frac{7381}{25200}\approx 0.29289683$$ while the above expansion gives $$\frac{59}{12000}+\frac{1}{10} (\gamma +\log (10))\approx 0.29289674$$

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I am not sure if I understood what you wrote. But I am quite sure that you do not need the measure you defined, or in fact any complicated calculus.

The function $f(x):= \sum\limits_{n=1}^{\infty} \frac{1}{n(n+x)}$ is clearly strictly monotone decreasing and non-negative. So the limit exists, and in fact $\lim\limits_{x\rightarrow \infty} f(x)$ is the limit of the sequence $\lim\limits_{m\rightarrow \infty} f(m)$ where $m$ runs through the positive integers.

So we can restrict our attention to positive integers $x$. Then $\frac{1}{n(n+x)} = \frac{1}{x} (\frac{1}{n}- \frac{1}{n+x})$. The sum $\sum\limits_{n=1}^{\infty} \frac{1}{n}- \frac{1}{n+x}$ is a telescopic sum, its value is $\sum\limits_{n=1}^x \frac{1}{n} \approx \log x$. So $f(x)\approx \frac{\log x}{x}$, which tends to zero as $x$ tends to infinity.

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  • $\begingroup$ It's not my question. I don't want to compute this limit. I just want to make a comparaison with measure theory. $\endgroup$ – user380364 Jul 22 '18 at 13:28
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    $\begingroup$ I do not think it can be effectively used to solve this problem. If anything, it seems to complicate matters rather than simplify them. $\endgroup$ – A. Pongrácz Jul 22 '18 at 13:30
  • $\begingroup$ On the contrary, it is a very simple approach to see that the terms converge to zero and then go looking for a dominant sequence and then fill in the issue of the real limit by monotonicity. DCT is often taught only later on, but the concept is simply enough "you are actually allowed to do what you have always wanted to do if you check this condition". $\endgroup$ – Phira Jul 22 '18 at 13:39
  • $\begingroup$ In fact the calculus you use here is much more complicated than necessary. $\endgroup$ – David C. Ullrich Jul 22 '18 at 14:03
  • $\begingroup$ We can make it more general without restricting $x$ to integers. Starting from your answer, I built mine. By the way, $+1$ for your to compensate a downvote which is not justified (at least for me). Cheers. $\endgroup$ – Claude Leibovici Jul 22 '18 at 14:11

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