The sum (from n=1 to infinity) of (n^2)(1/3)^n

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- May 6th 2010, 12:53 PMfeyomiCompute the following summation please(!)
The sum (from n=1 to infinity) of (n^2)(1/3)^n

Thanks - May 6th 2010, 02:13 PMSoroban
Hello, feyomi!

Quote:

$\displaystyle \text{Evaluate: }\;S \;=\;\sum^{\infty}_{n=1} \frac{n^2}{3^n}$

$\displaystyle \begin{array}{cccccccccc}

\text{We have:} & S &=& \dfrac{1}{3} + \dfrac{4}{3^2} + \dfrac{9}{3^3} + \dfrac{16}{3^4} + \dfrac{25}{3^5} + \dfrac{36}{3^6} + \hdots \\ \\[-2mm]

\text{Multiply by }\frac{1}{3}\!: & \dfrac{1}{3}S &=& \quad\;\; \dfrac{1}{3^2} + \dfrac{4}{3^3} + \dfrac{9}{3^4} + \dfrac{16}{3^5} + \dfrac{25}{3^6} + \hdots \\\end{array}$

$\displaystyle \begin{array}{ccccccc}\text{Subtract: } & \qquad \dfrac{2}{3}S &=& \dfrac{1}{3} + \dfrac{3}{3^2} + \dfrac{5}{3^3} + \dfrac{7}{3^4} + \dfrac{9}{3^5} + \dfrac{11}{3^6} + \hdots \end{array}$

$\displaystyle \begin{array}{cccccc}\text{Multiply by }\frac{1}{3}\!: & \dfrac{2}{9}S &=& \qquad \dfrac{1}{3^2} + \dfrac{3}{3^3} + \dfrac{5}{3^4} + \dfrac{7}{3^5} + \dfrac{9}{3^6} = \hdots \\

\end{array}$

$\displaystyle \begin{array}{ccccccc}\text{Subtract:} & \qquad\;\dfrac{4}{9}S &=& \dfrac{1}{3} + \dfrac{2}{3^2} + \dfrac{2}{3^3} + \dfrac{2}{3^4} + \dfrac{2}{3^5} + \dfrac{2}{3^6} + \hdots \end{array}$

$\displaystyle \text{We have: }\;\frac{4}{9}S \;=\;\frac{1}{3} + \frac{2}{3}\underbrace{\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \hdots \right)}_{\text{geometric series}} $

. . The geometric series has the sum: .$\displaystyle \frac{1}{1-\frac{1}{3}} \:=\:\frac{3}{2}$

Hence: . $\displaystyle \frac{4}{9}S \;=\;\frac{1}{3} + \frac{2}{3}\left(\frac{3}{2}\right) \;=\;\frac{1}{3} + 1 \quad\Rightarrow\quad \frac{4}{9}S\;=\;\frac{4}{3}$

Therefore: . $\displaystyle S \;=\;\frac{9}{4}\cdot\frac{4}{3} \quad\Rightarrow\quad \boxed{S\;=\;3}$

- May 7th 2010, 06:12 AMkompik
I think that another possible method for this problem would be summation by parts (I did not try myself), but Soroban's approach is very nice.

- May 7th 2010, 10:09 AMDrexel28
- May 7th 2010, 10:27 AMgmatt
I think that sort of argument requires uniform convergence so that you can swap the limit (there is an implicit limit when you take the derivative) unless you choose to treat things as formal power series, though then the question of convergence can't really be answered.

- May 7th 2010, 10:29 AMDrexel28
- May 7th 2010, 11:48 AMgmatt