I would like to calculate the sum for $x\in(-1,1)$ of this: $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$ So far I managed that $$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$ and $$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$ and $$ \int -\log(1-x) dx \overset{c}{=} x+(1-x)\log(1-x). $$ But I really don't know if I can put it somehow together or not.
Asked
Active
Viewed 109 times
4
-
Wolfram agrees with the result – kingW3 May 21 '17 at 14:26
-
It's the correct result. You just have to check the series and your answer have the same value if $x=0$. – Bernard May 21 '17 at 14:33
6 Answers
2
Yes, your answer is correct.
You can also see this by writing $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$.
Thomas Andrews
- 177,126
2
Let $f (x ) $ be your sum.
then
$$f'(x)=\sum_{n=1}^{+\infty}\frac {x^n}{n} $$ and $$f''(x)=\sum_{n=0}^{+\infty}x^n=\frac {1}{1-x} $$ thus $$f'(x)=-\ln (1-x) $$ and $$f (x)=x+(1-x)\ln (1-x) $$
hamam_Abdallah
- 62,951
2
Using partial fraction decomposition $$\frac 1{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ So $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty} \frac{x^{n+1}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x\sum_{n=1}^{\infty} \frac{x^{n}}n-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=x \log(1-x)-\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$$ Now, $$\frac d{dx} \left(\frac{x^{n+1}}{n+1}\right)=x^n\implies \frac d{dx} \left(\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} \right) =\frac{x}{1-x}=\frac{x-1+1}{1-x}=\frac{1}{1-x}-1$$ Now, integrate to get $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}=-\log(1-x)-x$$
Claude Leibovici
- 260,315
-
1I would personally do this: $\sum_{n=1}^\infty \frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty \left (\frac{x^{n}}{n} \right )-x = -\log(1-x)-x$ – Juan Carlos May 21 '17 at 14:51
-
@JuanCarlos. There are so many ways to do it ! I probably took a too long one. Cheers. – Claude Leibovici May 21 '17 at 14:55
1
Hint. One may write, for $|x|<1$, $$ \sum_{n=1}^\infty \frac{x^{n+1}}{n(n+1)}=x\sum_{n=1}^\infty \frac{x^n}{n}-\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)}=x+(1-x)\ln(1-x) $$ using $$ \sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)}=\sum_{m=1}^\infty \frac{x^m}{m}-\frac{x^1}{1}=-\ln(1-x)-x. $$
Olivier Oloa
- 120,989
1
When $|x| < 1$:
$$\begin{align*}\sum_{n=1}^\infty \dfrac{x^{n+1}}{n(n+1)} &= \sum_{n=1}^\infty \dfrac{x^{n+1}}{n} - \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^n}{n} + x \\ & = -x\ln(x-1) + \ln(x-1) + x\\ &= x+(1-x)\ln (1-x)\end{align*}$$
Darth Geek
- 12,296
0
Let $S(x)=\sum\limits_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}$, then on $x\in(-1,1)$, $$S''(x)=\left[\sum\limits_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}\right]'' =\sum\limits_{n=1}^{\infty}\left[\frac{x^{n+1}}{n(n+1)}\right]'' =\sum\limits_{n=1}^{\infty}{x^{n-1}}=\frac1{1-x},$$ hence, $$S(x)=\int\int\frac1{1-x}dxdx=-\int[\ln(1-x)+C_1]dx=x\ln(1-x)+(C_1-1)x-\ln(1-x)+C_2.$$ Let $x=0$ in the definition of $S(x)$, we can obtain $S(0)=0$ and $S'(0)=0$, then we know $C_1=C_2=0$, so $$S(x)=x\ln(1-x)-x-\ln(1-x).$$
ybtang21c
- 451