$$ \sum _{n=0}^{\infty }\:\frac{\left(n+1\right)\left(2n+1\right)}{3^n} $$
The sum should be $\frac{27}{4}$, but how do you calculate it ? Can someone give me an approach on how to do it?
thanks!
Find the sum of $\sum _{n=0}^{\infty }\:\frac{\left(n+1\right)\left(2n+1\right)}{3^n}$
2
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real-analysis
sequences-and-series
summation
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0you can try it with partial sums – 2017-01-29
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1or do not, just as well. – 2017-01-29
2 Answers
1
Consider that the power series $$ \sum_{n\geq 0}\frac{z^{2n+2}}{3^n} = \frac{3z^2}{3-z^2} $$ has a radius of convergence equal to $\rho=\sqrt{3}$. This allows us to differentiate both sides twice, then evaluate them at $z=1$, getting: $$ \sum_{n\geq 0}\frac{(2n+2)(2n+1)}{3^n} = \left.\frac{54(1+z^2)}{(3-z^2)^3}\right|_{z=1}=\frac{27}{2}. $$ Now it is enough to divide by $2$ both sides.
0
Hint $$\sum _{n=0}^{\infty }\:\frac{\left(n+1\right)\left(2n+1\right)}{3^n}=2\sum _{n=0}^{\infty }\frac{n^2}{3^n}+3\sum _{n=0}^{\infty }\frac{n}{3^n}+\sum _{n=0}^{\infty }\frac{1}{3^n}$$ On the other hand, if $|x|<1$, then $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$ as a result $$\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$$ and $$\sum_{n=2}^{\infty}n(n-1)x^{n-2}=\frac{2}{(1-x)^3}$$