3
$\begingroup$

How to find the limit $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}$ if we don't evaluate the sum?
I know the sum is actually an elementary function which we can find it using Fourier series or other methods, but I'm just curious about if there exists some alternative ways to find this limit.

I tried to write it as this form:

$$\displaystyle\lim_{x\rightarrow\infty}x\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right).$$

As we know, $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n\right)^{2}+x^{2}}$ and $\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x}{\left(2n-1\right)^{2}+x^{2}}$ must get a same value (we don't need to care about what the exact value is) , so this is in the form $``0\cdot\infty"$, which cannot be evaluated directly. This is where I get stucked.

After days of thinking, I'm getting closer to the answer. We can use easy algebra to get that $$\left |\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right |\leq\frac{1}{n^2}\quad\forall n\in\mathbb{Z^+},x\in\mathbb{R}$$
Hence the series below converges uniformly on $\mathbb{R}$:$$\displaystyle\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)$$
Changing the order of sum and limit, we can get:$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)=0$$
which is$$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$$
and we also know $$\displaystyle\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)=\lim_{x\rightarrow\infty}\left(-\frac{x^2}{1+x^2}-\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)\right)$$ If the limit exists, there must be an equation for the limit $L=-1-L$ which solves $L=-1/2$.
So everything needed is to prove that the limit exists. This would require a bit of analysis. I’m going to prove it via Cauchy’s rule ($\displaystyle\lim_{x\rightarrow+\infty}f\left(x\right)\ exists\Leftrightarrow\forall\epsilon>0\exists X>0 \forall x_1,x_2>X, \left|f(x_1)-f(x_2)\right|<\epsilon$).

$\endgroup$
  • $\begingroup$ @Masacroso I mean, probably you could evaluate this limit without knowing which function the series (which is taken limit) converges to. $\endgroup$ – Antimonius Dec 9 '17 at 1:21
  • $\begingroup$ Do you try by factoring out x^2 and canceling it? $\endgroup$ – Edumaths555 Dec 9 '17 at 1:42
  • $\begingroup$ @Edumaths555 I tried but nothing works — the order of “lim” and “Σ” can’t be changed. $\endgroup$ – Antimonius Dec 9 '17 at 1:51
  • $\begingroup$ Oh ya! need to apply Doninated Convergence Theorem or Fatou Lemma to change the order of lim and sum but still looking for proper setting. $\endgroup$ – Edumaths555 Dec 9 '17 at 2:01
  • $\begingroup$ @WongAustin Concering your recent edit - stuff like \dfrac and \displaystyle should be avoided in titles. (Where needed, you can use \limits command.) For more details see: Guidelines for good use of $\LaTeX$ in question titles. $\endgroup$ – Martin Sleziak Dec 9 '17 at 10:24
4
$\begingroup$

This answer tries to get things straight, but yes, there's a tiny piece missing in step $(4)$.


$\qquad(1)$: $\forall x \in\mathbb R$, the series $\sum_{n\geq 1}(-1)^n\frac{x^2}{n^2+x^2}$ converges absolutely.

Proof: We have that

$$\sum_{n\geq 1}\frac{x^2}{n^2+x^2}=x^2\,\sum_{n\geq 1}\frac{1}{n^2+x^2}\leq x^2\sum_{n\geq 1}\frac{1}{n^2}=\frac{x^2\pi^2}6.\qquad\qquad\square$$


$\qquad(2)$: $\forall x \in \mathbb R$, the series $\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right)$ converges and we have $x\sum_{n=1}^{\infty}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right)=\sum_{n\geq 1}(-1)^n\frac{x^2}{n^2+x^2}$.

Proof: Consider the partial sums

$$S_m=\sum_{n=1}^m(-1)^n\frac{x^2}{n^2+x^2}$$

and $$T_m=x\sum_{n=1}^{m}\left(\frac{x}{\left(2n\right)^{2}+x^{2}}-\frac{x}{\left(2n-1\right)^{2}+x^{2}}\right).$$

By $(1)$, $S_m$ converges as $m\to\infty$. It suffices to note that $T_m=S_{2m}$. $\square$


$\qquad(3)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)=0$

Proof: Expand

$$\pm\left (\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right )-\frac1{n^2}$$

and verify that the result is negative for all real $x$ and positive integers $n$. Conclude that the following estimate holds:

$$\left |\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right |\leq\frac{1}{n^2}\quad\forall n\in\mathbb{Z^+},\forall x\in\mathbb{R}$$

It then follows from the Weierstrass M-test that $\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)$ converges uniformly and absolutely on $\mathbb{R}$. Since uniform convergence holds, we have

\begin{align} &\lim_{x\to\infty}\sum_{n=1}^{\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)\\ =&\sum_{n=1}^{\infty}\lim_{x\to\infty}\left(\frac{2x^2}{\left(2n\right)^2+x^2}-\frac{x^2}{\left(2n-1\right)^2+x^2}-\frac{x^2}{\left(2n+1\right)^2+x^2}\right)\\ =&\sum_{n=1}^{\infty}(2-1-1)=\sum_{n=1}^{\infty}0=0 \end{align}

which concludes the proof. $\square$


$\qquad(4)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)$ and $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$ both exist, and they are equal.

Partial Proof: It follows from $(3)$ and the algebra of limits that

$$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right) = \lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right)$$

provided both limits exist.


$\qquad(5)$: $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}=-1/2$

Proof: For each $x\in\mathbb{R}$ we have

$$\sum_{n=1}^{m}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)\\ = -\frac{x^2}{1+x^2} -\sum_{n=1}^{m}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right) -\frac{x^2}{\left(2m+1\right)^{2}+x^2}.$$

Letting $m\to\infty$, we conclude that for all $x\in\mathbb{R}$

$$\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right) = -\frac{x^2}{1+x^2} -\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n+1\right)^{2}+x^{2}}-\frac{x^2}{\left(2n\right)^{2}+x^{2}}\right),$$

where the series on the LHS converges by $(2)$, and similarly the RHS series also converges.

Now, let $L=\lim_{x\to\infty}\sum_{n=1}^{\infty}\left(\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}\right)$. Letting $x\to\infty$ in the equality above and applying $(4)$, we get

$$L=-1-L\iff L=-1/2.$$

The claim follows from noting that $\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{2}}{n^{2}+x^{2}}=L$ as per $(2)$. $\square$


EDIT: We can use the integral test to arrive at the answer straight after step $(2)$. Indeed, for each $x\in\mathbb{R}$ let $a_x(n)=\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}$, so we are interested in $\lim_{x\to\infty}\sum_{n\geq 1}a_x(n)$.

Observe that $a_x(n)<0$ whenever $x\neq 0$ and $n\geq1$, so we may apply the integral test to $\sum_{n\geq 1}-a_x(n)$. We will have that

$$\int_1^\infty-a_x(t)\,dt\leq\sum_{n=1}^{\infty}-a_x(n)\leq -a_x(1) + \int_1^\infty-a_x(t)\,dt.$$

On the one hand, $-a_x(1)=\frac{x^2}{1+x^{2}}-\frac{x^2}{4+x^{2}}$. On the other,

\begin{align} \int_1^\infty-a_x(t)\,dt &=\int_1^\infty\frac{x^2}{\left(2t-1\right)^{2}+x^{2}}-\frac{x^2}{\left(2t\right)^{2}+x^{2}}\,dt \\&=-\frac{x}{2}\cdot\left[\arctan\left(\frac{1-2t}x\right)+\arctan\left(\frac{2t}x\right)\right]_{t=1}^\infty \end{align}

The brackets are simply $\left[\lim_{t\to\infty}\left(\arctan\left(\frac{1-2t}x\right)+\arctan\left(\frac{2t}x\right)\right)-\arctan\left(\frac{-1}x\right)-\arctan\left(\frac2x\right)\right]$, and since we have $\lim_{t\to\infty}\arctan\left(\frac{1-2t}x\right)=-\pi/2$ and $\lim_{t\to\infty}\arctan\left(\frac{2t}x\right)=\pi/2$, it follows that

$$\int_1^\infty-a_x(t)\,dt=\frac{x}{2}\left(\arctan\left(\frac{-1}x\right)+\arctan\left(\frac2x\right)\right)$$

Now, $\lim_{x\to\infty}-a_x(1)=0$ and $\lim_{x\to\infty}\int_1^\infty-a_x(t)\,dt=1/2$. This latter limit is easily computed considering the expansion

$$\arctan(z)=z-\frac{z^3}3+\frac{z^5}5-\dots$$

It follows from the squeeze theorem that $\sum_{n\geq 1}a_x(n)=-1/2$, which complets the proof. $\square$

$\endgroup$
  • $\begingroup$ Thanks a lot! I have got my final answer. Could you have a look at my answer below? $\endgroup$ – Antimonius Dec 11 '17 at 8:58
  • $\begingroup$ Very nice work after EDIT (+1). Unsurprisingly, its the same as the Abel summation of the divergent series $\sum_{n \geqslant1}^{\mathcal{A}} (-1)^n = \lim_{x \to 1-} \sum_{n=1}^\infty (-1)^n x^n$, although I wonder how that can be generalized beyond the power series. $\endgroup$ – RRL Dec 12 '17 at 23:18
0
$\begingroup$

Thanks for @Fimpellizieri , but I think I have got the answer.
We know $$\frac{x^2}{\left(2n\right)^{2}+x^{2}}-\frac{x^2}{\left(2n-1\right)^{2}+x^{2}}=\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}$$ Consider $$\left|\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right|\leq\frac{1}{n^2},\ \forall x\in\mathbb{R},n\in\mathbb{Z^+}$$ By Weierstrass M test the series below converges uniformly on $\mathbb{R}$. We have $$\lim_{x\rightarrow\infty}\sum_{n=1}^{\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]\\=\sum_{n=1}^{\infty}\lim_{x\rightarrow\infty}\left[\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}-\frac{\left(1-4n\right)x^2}{\left(\left(2n\right)^2+x^2\right)\left(\left(2n-1\right)^2+x^2\right)}\right]=0.$$
So we just need to consider $\displaystyle\sum_{n=1}^{\infty}\frac{-4nx^2}{\left(\left(2n\right)^2+x^2\right)^2}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{\frac{-4n}{x}}{\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$.
Then consider the "Riemann sum" of function $f\left(u\right)=\frac{-4u}{\left(4u^2+1\right)^2}$ on $\left[0,\infty\right)$ , we know the limit of the sum as $x\rightarrow\infty$ is actually the integral $$\int_{0}^{\infty}f\left(u\right)du=\frac{1}{8u^2+2}|_{0}^{\infty}=-\frac{1}{2}$$
which is the final answer.

Now I explain why the sum tends to the improper integral.
Notice that $f\left(u\right)$ is decreasing on $\left[0,\frac{1}{2}\right]$ and increasing on $\left[\frac{1}{2},+\infty\right)$, we split the sum into two parts $\displaystyle\sum_{0<2n\leq x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ and $\displaystyle\sum_{2n>x}\frac{\frac{-4n}{x}}{x\left(\left(\frac{2n}{x}\right)^2+1\right)^2}$ , noted respectively as $S_{1}\left(x\right)$ and $S_{2}\left(x\right)$.
$S_{1}\left(x\right)$ is obviously tending to the integral of $f\left(u\right)$ on $\left[0,\frac{1}{2}\right]$ , while $S_{2}\left(x\right)$ is bounded between $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+1\right)}^{\infty}f\left(u\right)du$ and $\displaystyle\int_{\frac{1}{x}\left(\left[\frac{x}{2}\right]+2\right)}^{\infty}f\left(u\right)du$ (due to monotonicity) , which both tend to $\displaystyle\int_{\frac{1}{2}}^{\infty}f\left(u\right)du$ .

$\endgroup$
  • $\begingroup$ I don't quite recognize that series as a Riemann sum. The Riemann sum of $f$ on $[0,x]$ looks something like $$\lim_{k\to\infty}\sum_{n=1}^{k}f\left(n\,\frac{x}k\right)\cdot\frac{x}k$$ $\endgroup$ – Fimpellizieri Dec 11 '17 at 13:41
  • $\begingroup$ Using the integral test, we can arrive at the answer easily after step $(2)$. I have added an edit to my answer with more details. $\endgroup$ – Fimpellizieri Dec 11 '17 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.