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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.

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The lower summation limit should probably be $n=1$, instead of $n=0$. – Sasha Dec 13 '11 at 1:59
HINT: $\int_0^x y^{n-1} \mathrm{d} y = \frac{x^n}{n}$. – Sasha Dec 13 '11 at 2:01
Your summand is $\frac{x^n}{n}$, no? It looks a bit like the integral of the geometric series... – J. M. Dec 13 '11 at 2:02
In your example do you mean $\sum \frac{3^n}{n!}$? – Qiaochu Yuan Dec 13 '11 at 2:02
$\sum_{n=1}^\infty \frac{1}{n3^n}$ is an expression, not a function. There is no variable. – Brandon Carter Dec 13 '11 at 2:06
up vote 9 down vote accepted

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


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

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

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


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

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 : $-\ln(1-\frac{1}{3})=\ln(\frac{3}{2})$

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