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$$ \sum_{n=0}^\infty \frac{x^n}{n(n+1)} $$ what is the sum of this? I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there

Michael Hardy
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Nestro Abarca
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    Is the $n(n+1)$ all in the denominator? Or do you mean $\frac{x^n}{n}\cdot(n+1)$? In the former, try partial fraction decomposition. Also, you start the sum from $n=0$ with $n$ in the denominator... – Clayton Sep 30 '15 at 03:09

5 Answers5

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$$ \sum\limits_{n=0}^{\infty}x^n=\frac1{1-x} $$ Integrate once to obtain $$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log\left({1-x}\right)$$ Integrate again to obtain $$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+1)(n+2)}=\left( 1-x \right)\log \left( 1-x \right) +x$$ $$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}=\left( 1-x \right)\log \left( 1-x \right) +x$$ $$\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}=\frac{\left( 1-x \right)\log \left( 1-x \right)}x +1$$

Pauly B
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You start with $$\sum_{n=0}^{\infty} x^n=\frac 1{1-x}$$ Now if you integrate it you get $$\int \sum_{n=0}^{\infty} x^n\;dx=\sum_{n=0}^{\infty} x^{n+1} / (n+1)=\int \frac 1{1-x}\; dx$$ You have to deal with the $n$ in the denominator-this should be a clue.

Ross Millikan
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Let $f(x)=\sum_{n=0}^\infty\frac{x^n}{n(n+1)}$ for $|x|<1$. Inside its radius of convergence we may differentiate a power series term by term, so $$f'(x)=\sum_{n=1}^\infty\frac{x^{n-1}}{n+1}=\frac1{x^2}\sum_{n=2}^\infty\frac{x^n}{n}=\frac{\log(1-x)-1}{x^2}.$$ Integrating, we have $$f(x)=f(0)+\int_0^x\frac{\log(1-t)-1}{t^2}\,\mathrm dt$$ and so you can calculate $f$ using integration by parts.

Jason
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Your sum has an error in it because for n=0 you are dividing by 0. However, if you start instead with n=1, there is no error, so I will assume you meant that.

Notice that $$\int \int x^n dx dx = \frac{x^{n+2}}{(n+1)(n+2)} +Cx+D =x \cdot \frac{x^{n+1}}{(n+1)(n+2)} +Cx+D$$

Therefore $$\int \int \sum_{n=0}^{\infty} x^n dx dx = Cx+D+ \sum_{n=0}^{\infty} x \cdot \frac{x^{n+1}}{(n+1)(n+2)} =Cx+D+ x \cdot \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)(n+2)}$$

$$ =Cx+D+ x \cdot \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} $$.

Letting $C = D = 0$ and dividing by $x$ gives a closed form expression for the sum. We also have that first integral, $$\int \int \sum_{n=0}^{\infty} x^n dx dx = \int \int \frac{1}{1-x} dx dx =\int C + \log(x-1) dx = Cx + D + (x-1) \log(x-1) $$

Again letting $C = D = 0$ and dividing by $x$, we have $$ \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \frac{(x-1)\log(x-1)}{x} $$

JuanTutors
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(Sum should be from $n=1$.)

$$\frac1{n(n+1)} = \frac1{n}-\frac1{n+1} $$

so that

$$\begin{align}\sum_{n=1}^{\infty} \frac{x^n}{n (n+1)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1} \\ &= -\log{(1-x)} - \frac1{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}\\ &=-\log{(1-x)} - \frac1{x} \left [-\log{(1-x)}-x \right ]\\ &= 1+\left ( \frac1{x}-1\right )\log{(1-x)} \end{align}$$

Note that the radius of convergence is $1$, i.e., the sum only converges when $|x| \le 1$.

Ron Gordon
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