# Compute the following summation please(!)

• May 6th 2010, 12:53 PM
feyomi
The sum (from n=1 to infinity) of (n^2)(1/3)^n

Thanks
• May 6th 2010, 02:13 PM
Soroban
Hello, feyomi!

Quote:

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

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

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

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

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

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

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

• May 7th 2010, 06:12 AM
kompik
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 AM
Drexel28
Quote:

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 $\sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''$
• May 7th 2010, 10:27 AM
gmatt
Quote:

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.
• May 7th 2010, 10:29 AM
Drexel28
Quote:

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
• May 7th 2010, 11:48 AM
gmatt
Quote:

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.