Compute the following summation please(!)

**feyomi** · May 6th 2010, 12:53 PM

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

Thanks

**Soroban** · May 6th 2010, 02:13 PM

Hello, feyomi!

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

$\begin{array}{cccccccccc}<br /> \text{We have:} & S &=& \dfrac{1}{3} + \dfrac{4}{3^2} + \dfrac{9}{3^3} + \dfrac{16}{3^4} + \dfrac{25}{3^5} + \dfrac{36}{3^6} + \hdots \\ \\[-2mm]<br /> \text{Multiply by }\frac{1}{3}\!: & \dfrac{1}{3}S &=& \quad\;\; \dfrac{1}{3^2} + \dfrac{4}{3^3} + \dfrac{9}{3^4} + \dfrac{16}{3^5} + \dfrac{25}{3^6} + \hdots \\\end{array}$

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

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

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

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

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

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

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

**kompik** · May 7th 2010, 06:12 AM

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.

**Drexel28** · May 7th 2010, 10:09 AM

Originally Posted by feyomi

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

Thanks

The other classical maneuver is to notice that $\sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''$

**gmatt** · May 7th 2010, 10:27 AM

Originally Posted by Drexel28

The other classical maneuver is to notice that $\sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''$

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.

**Drexel28** · May 7th 2010, 10:29 AM

Originally Posted by gmatt

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.

power series converge unilformly on their disk of convergence

**gmatt** · May 7th 2010, 11:48 AM

Originally Posted by Drexel28

power series converge unilformly on their disk of convergence

Yeap thats right, just pointing out the trick is being used implicitly.

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