another sum of the series

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June 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.
June 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 $S=\sum \frac{n+1}{3^n}$
Find 3S
Find 3S-S

KeepSmiling
Malay
June 25th 2006, 03:43 PM
Nichelle14

Can you explain a little more? Still lost :confused:
June 25th 2006, 08:13 PM
Soroban

Hello, Nichelle14!

Quote:

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

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

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

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

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

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

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