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I'd like to turn this sum: \begin{align}\sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)} \end{align} into an integral $\displaystyle \int_{a}^{b} g(x) \space dx$.

There seems to be many methods to either change or approximate sums as integrals. So I've become confused which approach would work.

In Is it possible to write a sum as an integral to solve it? robjohn used $\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n$ which looks similar to a Laplace Transforms.

I can't see how he gets rid of the n's so I'm not able to apply it here otherwise it seems promising. But looking elsewhere there are also approximations methods such as: Turning infinite sum into integral which even more obscure at least to me.

How do I convert this sum to an integral?

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    $\begingroup$ He got rid of the $n$'s by pulling the sum inside the integral. The sum with the exponentials was an easier problem than the original sum. $\endgroup$ – John Dec 19 '17 at 20:28
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Well, you could write $$\frac{1}{n+1} = \int_0^1 t^n\; dt$$ so (for $|x| < 3$) your sum becomes $$ \eqalign{\sum_{n=0}^\infty &\left(\frac{x}{3}\right)^{n+1} \int_0^1 t^n\; dt\cr = & \frac{x}{3} \int_0^1 \sum_{n=0}^\infty \left(\frac{xt}{3}\right)^n \; dt\cr = & \frac{x}{3} \int_0^1 \frac{dt}{1-xt/3}\cr = & \ln\left(\frac{3}{3-x}\right)\cr }$$

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Recall that $$\frac{x^{n+1}}{3^{n+1}(n+1)} = \int_0^{x/3} t^n \ \mathrm{d}t$$ Hence $$\sum_{n=0}^{\infty} \frac{x^{n+1}}{3^{n+1}(n+1)} = \sum_{n=0}^{\infty} \int_0^{x/3} t^n \ \mathrm{d}t = \int_0^{x/3} \sum_{n=0}^{\infty} t^n \ \mathrm{d}t = \int_0^{x/3} \frac{1}{1-t} \mathrm{d}t = -\ln \left( \frac{3-x}{3} \right)$$

This holds for $|x|<3$.

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If you differentiate it you will get a geometric series $$\begin{align} \frac d{dx}\sum_{n=0}^\infty \frac {x^{n+1}}{3^{n+1}(n+1)} &=\sum_{n=0}^\infty\frac d{dx}\frac 1{(n+1)}\left(\frac x3\right)^{n+1}\\ &=\sum_{n=0}^\infty\left(\frac x3\right)^n\\ &=\frac 1{1-\frac x3}\\ &=\frac 3{3-x}\\ \text{Integrating:} \qquad \sum_{n=0}^\infty \frac {x^{n+1}}{3^{n+1}(n+1)}&=-\ln\left(\frac {3-x}3\right)=\ln\left(\frac 3{3-x}\right)\end{align}$$

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* by Fubini/Tonelli's theorem

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