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I need to know what the function $\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$ represents evaluated at a particular point.

For example if the series given was $\sum \limits_{n=0}^{\infty}\frac{3^n}{n!}$ the answer would be $e^x$ evaluated at $3$.

Yes, this is homework, but I'm not looking for any handouts, any help would be greatly appreciated.

  • 6
    $\sum_{n=1}^\infty \frac{1}{n3^n}$ is an expression, not a function. There is no variable.2011-12-13

3 Answers 3

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Let $f(x)=\sum_1^{\infty}{1\over nx^n}$; you've got $f(3)$, so you want to know what $f(x)$ is. Differentiate it.

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Take $f(x) = \displaystyle \sum_{n=1}^\infty \frac{x^n}{n}$.

Then,

$f^\prime (x) = \displaystyle \sum_{n=1}^\infty \frac{n x^{n-1}}{n} = \sum_{n=0}^\infty x^n$.

The last expression is a geometric series and, as long as $x < 1$, it can be expressed as

$f^\prime (x) = \displaystyle \frac{1}{1-x}$.

Therefore,

$f(x) = - \ln | 1 - x | + \kappa$

Where $\kappa$ is a constant. But if you take the original expression for $f(x)$, you can see that $f(0) = 0$ and, therefore, $\kappa = 0$.

So $f(x) = -\ln | 1 - x |$.

The answer to your question is just $f \left(\frac{1}{3} \right)$.

You can also obtain this result by Taylor expanding $\ln ( 1 - x )$.

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We have $ \displaystyle -\ln(1-x) = \sum_{n \geq 1 } \frac{x^n}{n} $ by course for $|x|<1$
So this is simply $f(x)=-ln(1-x) $ evaluated at $x=\frac{1}{3}$
Which gives the value : $-\ln(1-\frac{1}{3})=\ln(\frac{3}{2})$