1. ## Compute the following summation please(!)

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

Thanks

2. Hello, feyomi!

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

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

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

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

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

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

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

3. 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.

4. 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 $\displaystyle \sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''$

5. Originally Posted by Drexel28
The other classical maneuver is to notice that $\displaystyle \sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''$
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.

6. 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

7. 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.