Compute $\lim\limits_{x\to \infty }\sum\limits_{n=1}^\infty \frac{1}{n(n+x)}$

Question

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 ?

Answer 1

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.

Answer 2

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.

Answer 3

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

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

Answer 4

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 $$\frac{59}{12000}+\frac{1}{10} (\gamma +\log (10))\approx 0.29289674$$