1. ## Sum of Series

Find the sum of the series $\displaystyle \sum _{n=3}^{\infty} \frac{n}{3^n}$

Thanks

2. ## Ok

Originally Posted by polymerase
Find the sum of the series $\displaystyle \sum _{n=3}^{\infty} \frac{n}{3^n}$

Thanks
I will explain later but the answer is $\displaystyle \frac{7}{36}$

3. Originally Posted by polymerase
Find the sum of the series $\displaystyle \sum _{n=3}^{\infty} \frac{n}{3^n}$

Thanks
For $\displaystyle |x|<1$:

$\displaystyle \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \implies \left( \sum_{n=0}^{\infty} x^n \right)' = \frac{1}{(1-x)^2} \implies \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$

Use that to solve the problem.

4. Hello, polymerase!

Find the sum of the series: .$\displaystyle \sum _{n=3}^{\infty} \frac{n}{3^n}$

We have: . . . . .$\displaystyle S \;=\;\frac{3}{3^3} + \frac{4}{3^4} + \frac{5}{3^5} + \frac{6}{3^6} + \cdots$

Multiply by $\displaystyle \frac{1}{3}\!:\;\;\frac{1}{3}S \;=\qquad\quad\frac{3}{3^4} + \frac{4}{3^5} + \frac{5}{3^6} + \hdots$

Subtract:. . . . .$\displaystyle \frac{2}{3}S \;=\;\frac{3}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \frac{1}{3^6} + \cdots$

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

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

So we have: . $\displaystyle \frac{2}{3}S \;=\;\frac{1}{9}+ \frac{1}{81}\!\left(\frac{3}{2}\right) \;=\;\frac{7}{54}$

Therefore: . $\displaystyle S \;=\;\frac{3}{2}\!\left(\frac{7}{54}\right) \;=\;\frac{7}{36}$

5. ## Just wondering

Originally Posted by ThePerfectHacker
For $\displaystyle |x|<1$:

$\displaystyle \sum_{n=0}^{\infty}x^n = \frac{1}{1-x} \implies \left( \sum_{n=0}^{\infty} x^n \right)' = \frac{1}{(1-x)^2} \implies \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$

Use that to solve the problem.
but for this case then would it be $\displaystyle \sum_{n=3}^{\infty}nx^{n}=\frac{x^3}{(1-x)^2}$