sequences and series - Can we find the limit $lim_{xrightarrowinfty}sum_{n=1}^{infty}frac{left(-1right)^nx^2}{n^2+x^2}$ without evaluating the sum?

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 ($\displaystyle\lim_{x\rightarrow+\infty}f\left(x\right)\ exists\Leftrightarrow\forall\epsilon>0\exists X>0 \forall x_1,x_2>X, \left|f(x_1)-f(x_2)\right|<\epsilon$).

Answer

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 $T_m=S_{2m}$. $\square$

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

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.

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

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. $\square$

limits - Prove $lim_{xtoinfty}frac{x^{x^2}}{2^{2^x}}$

How do you prove this formula

$$\lim_{x \to\infty}\frac{x^{x^2}}{2^{2^x}}$$

Since both top and bottom approaches infinity, I assume it is L'Hospital's rule to solve it, but after the first step I'm stuck
$$\lim_{x \to\infty}\frac{x^2 logx}{2^xlog2}$$
So how can I solve this problem, it seems the answer is infinity but I don't know how to approach that.

Answer

You can rewrite the limit as
\begin{aligned}
L&=\lim_{x \to\infty}\frac{x^{x^2}}{2^{2^x}}\\
&=\lim_{x \to\infty}\frac{\exp\left(x^2\cdot\log x\right)}{\exp\left(2^x\cdot\log 2\right)}\\

&=\lim_{x \to\infty}\exp\left(x^2\cdot\log x-2^x\cdot\log 2\right)
\end{aligned}

Since $\exp(.)$ is a continuous function, we can change the order to
$$L=\exp\left(\lim_{x\to\infty}\left(x^2\cdot\log x-2^x\cdot\log 2\right)\right)$$

and name the inner limit to $L'$
$$L'=\lim_{x\to\infty}\left(x^2\cdot\log x-2^x\cdot\log 2\right)$$

Note that $\log x

Obviously, the limit of the RHS expression as $x$ goes to $\infty$ is $-\infty$; because the growth of $x^3$ is not comparable with the exponential growth of $2^x$. Hence, $L'=-\infty$, which means $L=0$.

complex analysis - Show that $intnolimits^{infty}_{0} x^{-1} sin x dx = fracpi2$

Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ by integrating $z^{-1}e^{iz}$ around a closed contour $\Gamma$ consisting of two portions of the real axis, from -$R$ to -$\epsilon$ and from $\epsilon$ to $R$ (with $R > \epsilon > 0$) and two connecting semi-circular arcs in the upper half-plane, of respective radii $\epsilon$ and $R$. Then let $\epsilon \rightarrow 0$ and $R \rightarrow \infty$.

[Ref: R. Penrose, The Road to Reality: a complete guide to the laws of the universe (Vintage, 2005): Chap. 7, Prob. [7.5] (p. 129)]

Note: Marked as "Not to be taken lightly", (i.e. very hard!)

Update: correction: $z^{-1}e^{iz}$ (Ref: http://www.roadsolutions.ox.ac.uk/corrections.html)

calculus - Why "$limlimits_{xrightarrow infty} frac{x+sin x}{x}$ does not exist" is not an acceptable answer?

Find the limits:

$\lim\limits_{x\rightarrow \infty} \frac{x+\sin x}{x}$

Since the numerator and denominator tends to infinity as $x$ tends to infinity, then applying Lhopital's rule:

$\lim\limits_{x\rightarrow \infty} \frac{x+\sin x}{x} = \lim\limits_{x\rightarrow \infty} \frac{1 + \cos x}{1}$

since $\cos x$ has no limit as $x$ tend to infinity ($\cos x$ oscillates between $-1$ and $1$), I conclude that the limit of $\frac{x + \sin x }{x}$ as $x$ tends to infinity does not exist too.

Why is this answer wrong (the correct answer is 1) and at what point should I have realized that I made a mistake and abort this solution and try something else? Is $\frac{1 + \cos x}{1}$ considered an indeterminate form?

Answer

L'Hospital rule says that if $\lim_{x \to \infty} f(x)=\lim_{x \to \infty}g(x)= \infty$ and $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ exists then $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ exists and is the same thing.

L'Hospital doesn't say anything about what happens if $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ doesn't exists, and the converse of L'H is not true.

To understand why this happens, you have to look at the proof of L'H. The proof of L'H uses the mean value theorem to deduce that there exists some $c_x$ which depends on $x$ such that
$c_x \to \infty$ and
$$\frac{f(x)}{g(x)}= \frac{f'(c_x)}{g'(c_x)}$$

Now, $\lim_x \frac{f'(x)}{g'(x)}$ calculates the limit using ALL $x$, while $\lim_x\frac{f'(c_x)}{g'(c_x)}$ calculates the limit using only SOME of the $x$. If the limit exists for all $x$, then it does exists for some of the $x$, but the other way around it is not true.

In your example, what happens most probably is that while
$$\lim\limits_{x\rightarrow \infty} \frac{1 + \cos x}{1}$$
does not exist, when you apply the MVT you get

$$\frac{x+\sin x}{x} =\frac{1 + \cos c_x}{1}$$
and each $c_x$ is very very close to some $\frac{\pi}{2}+2k \pi$. Moreover, when $x$ goes to infinity the approximation $c_x \sim \frac{\pi}{2}+2k \pi$ becomes better and better.

matrices - Finding trace and determinant of a matrix

The trace and determinant of a 3x3 matrix satisfy Tr A=2 and det A=2. The sum of two eigenvalues of A is equal to the third eigenvalue. Then the trace and determinant of the matrix $A^2$ is equal to?

I know that the trace is equal to the sum of eigenvalues and determinant is equal to its products.

Let $\lambda_1,\lambda_2, \lambda_3$ be the eigenvalues

$\lambda_1+\lambda_2+\lambda_3= 2$

$\lambda_1\lambda_2\lambda_3=2$

$\lambda_1+\lambda_2=\lambda_3$

Even if i make some substitutions I do not know how to get it for $A^2$.

Please explain how to do this.

Answer

From $\lambda_1+\lambda_2+\lambda_3=2$ and $\lambda_1+\lambda_2-\lambda_3=0$, we obtain $\lambda_1+\lambda_2=\lambda_3=1$.

From $\lambda_1\lambda_2\lambda_3=2$ we obtain $\lambda_1\lambda_2=2$.

Therefore, $\lambda_1^2+\lambda_2^2=(\lambda_1+\lambda_2)^2-2\lambda_1\lambda_2=-3$.

Therefore, $\lambda_1^2+\lambda_2^2+\lambda_3^2=-2$.

---

If $A\vec v = \lambda \vec v$, then $A^2 \vec v = \lambda^2 \vec v$. Therefore, $\lambda^2$ are the eigenvalues of $A^2$.

Therefore, $\operatorname{tr}(A^2) = \lambda_1^2+\lambda_2^2+\lambda_3^2=-2$.

$\det(A^2) = \det(A)^2 = 2^2 = 4$.

complex numbers - Summation of $frac{cos n theta}{2^n}$

I would like to compute the following sum:

$$\sum_{n=0}^{\infty} \frac{\cos n\theta}{2^n}$$

I know that it involves using complex numbers, although I'm not sure how exactly I'm supposed to do so. I tried using the fact that $\cos \theta = {e^{i\theta} + e^{-i\theta}\over 2}$. I'm not sure how to proceed from there though. A hint would be appreciated.

Answer

Consider the series
$$S=\sum_{n=0}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^n.$$
This is a geometric series whose sum is
$$S=\frac{2}{2-e^{i\theta}}.$$

Now the real part of $S$ is the sum you are looking for.

calculus - $dfrac{2}{pi} = dfrac{sqrt 2}{2} cdot dfrac{sqrt {2+sqrt 2}}{2} cdotdfrac{sqrt {2+sqrt {2+sqrt 2}}}{2} cdots $

One can show inductively that
$$
\cos \frac{\pi}{2^{n+1}}\ = \frac{\sqrt {2+\sqrt {2+\sqrt {2+\sqrt {\cdots+\sqrt {2 }}}}}}{2},
$$
with $n$ square roots in the right side of the equation.

The second part of the question was to deduct the following from the first part:

$$\frac{2}{\pi} = \frac{\sqrt 2}{2} \cdot \frac{\sqrt {2+\sqrt 2}}{2} \cdot\frac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdot \cdots $$

with the hint to use the following limit:

$$\lim_{n\to \infty}\cos\Big(\frac{t}{2}\Big)\cos\Big(\frac{t}{2^2}\Big)\cdots\cos\Big(\frac{t}{2^n}\Big) = \frac{\sin t}{t}.$$

A hint or some general intuition will be appreciated.

Answer

Using the trogonometric identity
$$

\sin (2a)=2\sin a\,\cos a\qquad\text{or}\qquad \cos a=\frac{\sin 2a}{2\sin a},
$$
provided that $\,\sin a\ne 0,\,$ we obtain that
$$
\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\cdots\frac{\sin (x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin x}{2^n\sin(x/2^n)}.
$$
Hence

$$
\lim_{n\to\infty}\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{x},

$$
since
$$
\lim_{t\to 0}\frac{\sin(tx)}{t}=x.
$$
In particular,
$$
\prod_{n=1}^\infty \cos\left(\frac{\pi}{2^{n+1}}\right)=\frac{2}{\pi}.
$$