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I've been trying to find the coefficient of $x^n$ in the expansion of $ { ({\log_e (1+x) })^2 } $.I wrote out the expansion of $ { ({\log_e (1+x) })^2 } $ explicitly and tried to generalize the terms involving $x^n$, but...so far no luck.

Is there any other alternative ?

Diego Silvera
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donvoldy666
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2 Answers2

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Squaring $\log(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}...\ $ one obtains the following coefficient of $x^n$: $$\sum_{k=1}^{n-1}(-1)^{n}\dfrac{1}{k(n-k)}=\dfrac{(-1)^n}{n}\sum_{k=1}^{n-1}\dfrac{1}{k}+\dfrac{1}{n-k}=\dfrac{(-1)^n\cdot2H_{n-1}}{n}\ .$$

Christian Blatter
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Ishan Banerjee
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  • How did you perform this split $ \sum_{k=1}^{n-1}(-1)^{n}\dfrac{1}{k(n-k)}=\dfrac{(-1)^n}{n}\sum_{k=1}^{n-1}\dfrac{1}{k}+\dfrac{1}{n-k} $ ? – donvoldy666 Mar 23 '13 at 11:43
  • It's partial fractions.$(-1)^n$ and $ n$ doesn't depend on k so I could take them outside the summation. – Ishan Banerjee Mar 23 '13 at 11:45
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If $$y=\{\log(1+x)\}^2$$

So, $$\frac{dy}{dx}=2\frac{\log(1+x)}{(1+x)}=2 (1+x)^{-1} \log(1+x)$$ $$=2\left(1+\frac{x(-1)}{1!}+\frac{x^2(-1)(-2)}{2!}+\frac{x^3(-1)(-2)(-3)}{3!}+\cdots\right)$$ $$\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\right)$$

$$=2 (1-x+x^2-x^3+\cdots )\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots\right)$$

$$=2\{x+x^2\left(-\frac12-1\right)+x^3\left(\frac13+\frac12+1\right)+x^4\left(-\frac14-\frac13-\frac12-1\right)+\cdots\}$$

So, $$ dy =2\sum_{1\le r<\infty} H_rx^r(-1)^{r-1} dx$$ where $H_n=\sum_{1\le r\le n}\frac1r$

Integrating both sides, $y= 2\sum_{1\le r<\infty} \frac{H_rx^{r+1}(-1)^{r-1}}{r+1}+C$ where $C$ is an arbitrary constant for indefinite integral.

As at $x=0, y=\{\log(1+x)\}^2=0\implies C=0 $

$$\implies\{\log(1+x)\}^2=2\sum_{1\le r<\infty} \frac{H_rx^{r+1}(-1)^{r-1}}{r+1}$$

lab bhattacharjee
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