Compute the following sum.

**feyomi** · May 6th 2010, 12:16 PM

The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

Thanks.

**Anonymous1** · May 6th 2010, 12:29 PM

Originally Posted by feyomi

The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

Thanks.

$\sum_{i=1}^{\infty} (\frac{1}{3})^{n-1} = [\sum_{i=0}^{\infty} (\frac{1}{3})^{n-1}] - (\frac{1}{3})^{-1} = \frac{9}{2} - 3 = \frac{3}{2}$

Since,

$\sum_{i=0}^{\infty} (\frac{1}{3})^{n-1}=(\frac{1}{3})^{-1} \sum_{i=0}^{\infty} (\frac{1}{3})^{n} = \frac{(\frac{1}{3})^{-1}}{1 - \frac{1}{3}} = \frac{9}{2}$

**Archie Meade** · May 6th 2010, 12:39 PM

Originally Posted by feyomi

The sum (from n=1 to infinity) of ng(n), where g(n)=(1/3)^(n-1)

Thanks.

Hi feyomi,

$S_{\infty}=(1)\left(\frac{1}{3}\right)^0+(2)\left( \frac{1}{3}\right)^1+(3)\left(\frac{1}{3}\right)^2 +....$

$=(1)+(2)\left(\frac{1}{3}\right)^1+(3)\left(\frac{ 1}{3}\right)^2+....$

$\frac{1}{3}S_{\infty}=\frac{1}{3}+(2)\left(\frac{1 }{3}\right)^2+(3)\left(\frac{1}{3}\right)^3+....$

$S_{\infty}-\frac{1}{3}S_{\infty}=1+(2-1)\left(\frac{1}{3}\right)+(3-2)\left(\frac{1}{3}\right)^2+(4-3)\left(\frac{1}{3}\right)^3+.....$

$=1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\f rac{1}{3}\right)^3+....$

$=geometric\ series$

For the geometric series, the first term is 1 and the common ratio is 1/3,
hence it's sum to infinity is

$\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$

$\frac{2}{3}S_{\infty}=\frac{3}{2}$

$S_{\infty}=\frac{3}{2}\frac{3}{2}=\frac{9}{4}$

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