- #1
Kernul
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- 7
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?