# another sum of the series

• Jun 24th 2006, 07:54 PM
Nichelle14
another sum of the series
How do I find the sum of this series?

summation of (n+1)/3^n

when I used the first few n's I didn't see a pattern.
• Jun 24th 2006, 09:37 PM
malaygoel
Quote:

Originally Posted by Nichelle14
How do I find the sum of this series?

summation of (n+1)/3^n

when I used the first few n's I didn't see a pattern.

It is the combination of arithmetic and geometric series.
Let $\displaystyle S=\sum \frac{n+1}{3^n}$
Find 3S
Find 3S-S

KeepSmiling
Malay
• Jun 25th 2006, 03:43 PM
Nichelle14
Can you explain a little more? Still lost :confused:
• Jun 25th 2006, 08:13 PM
Soroban
Hello, Nichelle14!

Quote:

How do I find the sum of this series? .$\displaystyle \sum^{\infty}_{n=1} \frac{n+1}{3^n}$

We are given: . . $\displaystyle S \;= \;\frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \frac{5}{3^4} + \,\cdots$

Multiply by $\displaystyle \frac{1}{3}:\;\;\frac{1}{3}S \;= \qquad\;\frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \,\cdots$

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

After the first term, we have a geometric series with first term $\displaystyle a = \frac{1}{3^2}$
. . common ratio $\displaystyle r = \frac{1}{3}$, and whose sum is: $\displaystyle \frac{\frac{1}{3^2}}{1 - \frac{1}{3}}\:=\:\frac{1}{6}$

Hence, we have: .$\displaystyle \frac{2}{3}S\:=\:\frac{2}{3} + \frac{1}{6} \:=\:\frac{5}{6}$

Therefore: .$\displaystyle \boxed{S\:=\:\frac{5}{4}}$