Is the Summation Converging in the Given Interval?

[Kernul]

Homework Statement

I'm give the following summation of functions and I have to see where it converge.
$$\sum_{n = 1}^{\infty} \frac{(3 arcsin x)^n}{\pi^{n + 1}(\sqrt(n^2 + 1) + n^2 + 5)}$$

Homework Equations

The Attempt at a Solution

Putting $3 arcsin x = y$, I already see that with the theorem of D'Alambert I have a range $r = \pi$.
The summation then converges in the interval:
$$|3 arcsin x| < \pi$$
which is
$$sin(-\frac{\pi}{3}) < x < sin(\frac{\pi}{3})$$
Now I don't know it the summation converges in the two external points, so I try with the first one and I get:
$$\frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{(-1)^n}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
In this case, I decide to use the Leibniz criterion. So first I find if the summation converges to $0$ and then I find if decreases monotonically.
Doing the limit of the summation (minus the $(-1)^n$) I get that it converges to $0$.
Now, to find out if it decreases monotonically, I do the first derivative of the following:
$$f(n) = \frac{1}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
and I end up with
$$f'(n) = -\frac{\frac{x}{\sqrt(x^2 + 1)} + 2x}{(\sqrt(n^2 + 1) + n^2 + 5)^2}$$
Now I have to put this $>0$ but due to the minus sign it will be $<0$. The one at the denominator is always positive, so we have to look at only the numerator. We will have then:
$$\frac{x}{\sqrt(x^2 + 1)} + 2x < 0$$
But, strangely, I don't get how to continue from here.
Can someone help me?

 

[fresh_42]

$x \mapsto \frac{x}{\sqrt{x^2+1}}+2x$ is positive for $x>0$, negative for $x<0$ and zero at $x=0$.

 

[haruspex]

I do not understand how differentiating f(n) wrt n caused x to appear from nowhere.
Since 1/n² converges, it is rather obvious that the sum you have converges at both ends of the range. Is there not a test along those lines?

 

[Ray Vickson]

Homework Statement

I'm give the following summation of functions and I have to see where it converge.
$$\sum_{n = 1}^{\infty} \frac{(3 arcsin x)^n}{\pi^{n + 1}(\sqrt(n^2 + 1) + n^2 + 5)}$$

Homework Equations

The Attempt at a Solution

Putting $3 arcsin x = y$, I already see that with the theorem of D'Alambert I have a range $r = \pi$.
The summation then converges in the interval:
$$|3 arcsin x| < \pi$$
which is
$$sin(-\frac{\pi}{3}) < x < sin(\frac{\pi}{3})$$
Now I don't know it the summation converges in the two external points, so I try with the first one and I get:
$$\frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{(-1)^n}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
In this case, I decide to use the Leibniz criterion. So first I find if the summation converges to $0$ and then I find if decreases monotonically.
Doing the limit of the summation (minus the $(-1)^n$) I get that it converges to $0$.
Now, to find out if it decreases monotonically, I do the first derivative of the following:
$$f(n) = \frac{1}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
************
Can someone help me?

You have $f(n) = 1/d(n)$, where $d(n) = \sqrt{n^2+1} + n^2+5$ is a strictly increasing function of $n$. No calculus is needed here: $n^2$ is increasing in $n$, so $\sqrt{n^2+1}$ is increasing, as is $n^2+5$. You are just adding two increasing functions.

 

[Kernul]

$x \mapsto \frac{x}{\sqrt{x^2+1}}+2x$ is positive for $x>0$, negative for $x<0$ and zero at $x=0$.

Why? How did you do that?

I do not understand how differentiating f(n) wrt n caused x to appear from nowhere.
Since 1/n² converges, it is rather obvious that the sum you have converges at both ends of the range. Is there not a test along those lines?

Sorry, my bad! I meant "n" where there was "x".

You have $f(n) = 1/d(n)$, where $d(n) = \sqrt{n^2+1} + n^2+5$ is a strictly increasing function of $n$. No calculus is needed here: $n^2$ is increasing in $n$, so $\sqrt{n^2+1}$ is increasing, as is $n^2+5$. You are just adding two increasing functions.

So it decreases monotonically? This means that both this and the other one for $\frac{\pi}{3}$ converge and so it all converges to the interval $[-\frac{\pi}{3} , \frac{\pi}{3}]$?

 

[Ray Vickson]

Why? How did you do that?Sorry, my bad! I meant "n" where there was "x".So it decreases monotonically? This means that both this and the other one for $\frac{\pi}{3}$ converge and so it all converges to the interval $[-\frac{\pi}{3} , \frac{\pi}{3}]$?

Why do you assume you have convergence at both end points?

 

[fresh_42]

Why? How did you do that?

$\frac{x}{\sqrt{x^2+1}}+2x=x \cdot \underbrace{(2+\frac{1}{+\sqrt{x^2+1}})}_{>\,2}$