Finding the Sum of an Infinite Series: \sum_{0}^{\infty} \frac {n^2} {3^n}

wilcofan3
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Homework Statement



Find the sum \sum_{0}^{\infty} \frac {n^2} {3^n}

Homework Equations


The Attempt at a Solution



I don't know how to go about finding this sum, I have a guess of what it will be just by adding the first ten terms or so, but how do I find an actual approximation?
 
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Try multiplying three geometric series together and see how close it is to the above series. I think it's simmilar to repeated roots in differential equations.
 
John Creighto said:
Try multiplying three geometric series together and see how close it is to the above series. I think it's simmilar to repeated roots in differential equations.

The only similar series I see here are:

\sum_{0}^{\infty} (\frac {1} {3})^n

\sum_{0}^{\infty} n^2
 
Try looking at the second derivative of \sum_0^\infty x^n.
 
homogeneous differentiation

sum=[(xD)^2](1/(1-x))|x=1/3

that is
sum=g(1/3)
when
f(x)=1/(1-x)
and
g(x)=x[x*f'(x)]'=(x^2)*f''(x)+x*f'(x)
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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