1. Sum of Series

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

Thanks

2. Ok

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

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

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

Thanks
For $|x|<1$:

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

Use that to solve the problem.

4. Hello, polymerase!

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

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

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

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

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

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

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

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

5. Just wondering

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

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

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