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I m stuck in finding the sum $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n3^n}.$ Please give me some hint.

Raio
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    Integrate the Maclaurin expansion for $1/(1-x)$. I can add more detail in an answer below if necessary. Here is a similar problem for reference: http://math.stackexchange.com/questions/1706647/computing-the-series-sum-n-1-infty-frac1n2-2n/1706667#1706667 – User8128 Mar 22 '16 at 19:31

3 Answers3

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$$ \sum_{n=1}^\infty \frac{1}{n 3^n} = \sum_{n=1}^\infty \frac {x^n} n \quad \text{where } x = \frac 1 3. \tag 1 $$ \begin{align} \frac d {dx} \sum_{n=1}^\infty \frac{x^n} n \overset{\text{ ? }}{=} \sum_{n=1}^\infty \frac d {dx}\,\frac{x^n} n = \sum_{n=1}^\infty x^{n-1} = \frac 1 {1-x}. \end{align} So the sum is an antiderivative of $\dfrac 1 {1-x}$, evaluated at $x=1/3$. Which antiderivative of that function is it? It is the one that is equal to $0$ when $x=0$, because the expression in $(1)$ above is equal to $0$ when $x=0$.

PS suggested in "Dr. MV" 's comment: $\displaystyle \frac d {dx} \sum_{n=1}^\text{something} g_n(x) = \sum_{n=1}^\text{something} \frac d {dx} g_n(x)$ when $\text{“something''}$ (the number of terms being added) is finite. But this equality doesn't always hold when $\text{“something''}$ is $\infty$. That is why the $\text{“?''}$ surmounts the $\text{“}=\text{''}$ above. However it is true when the sum is a convergent power series, provided $x$ is in the interior (not on the boundary) of the disk of convergence.

Michael Hardy
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    Michael, solid answer; just a suggestion that might be useful to some readers ... consider explaining the reason that a question mark appears over the first "equals" sign in the second line. The validity of interchanging orders of operation (e.g., UC of the series obtained by differentiating term by term along with convergence of the original series at a point) is a key to the development. ;-)) - Mark – Mark Viola Mar 22 '16 at 21:50
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Here, we use a standard "trick" for evaluating certain types of series. The key of the development relies on the observation that

$$\int_0^x t^{n-1}\,dt=\frac{x^n}{n} \tag 1$$

Then, summing both sides of $(1)$ over $n$, we obtain for $|x|<1$

$$\begin{align} \sum_{n=1}^{\infty}\frac{x^n}{n}&=\sum_{n=1}^{\infty}\int_0^x t^{n-1}\,dt\\\\ &=\lim_{N\to \infty}\sum_{n=1}^N \int_0^x t^{n-1}\,dt\\\\ &=\lim_{N\to \infty} \int_0^x \sum_{n=1}^N t^{n-1}\,dt\\\\ &=\lim_{N\to \infty}\int_0^x\frac{1-t^N}{1-t}\,dt\\\\ &=\int_0^x \lim_{N\to \infty}\left(\frac{1-t^N}{1-t}\right)\,dt\\\\ &=\int_0^x \frac{1}{1-t}\,dt\\\\ &=-\log(1-x) \end{align}$$

where the Dominated Convergence Theorem guarantees the validity of the third to last equality. Now, set $x=1/3$.

Mark Viola
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$$\sum_{n = 1}^{+\infty} \frac{1}{n3^n} = \sum_{n = 1}^{+\infty}\frac{3^{-n}}{n}$$

This has the form of the well known series

$$\sum_{k = 1}^{+\infty}\frac{a^{-k}}{k} = - \log\left(\frac{a-1}{a}\right)$$

Thence

$$\sum_{n = 1}^{+\infty} \frac{1}{n3^n} = -\log\left(\frac{3-1}{3}\right) = - \log\frac{2}{3} = \log\frac{3}{2}$$

Enrico M.
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