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My partner is tutoring a Civil Engineering student in maths. Conveniently, I am a civil engineer so when any maths that's confusing comes up, I can usually help out. However, we are having a problem with the summing of power series, one area where I am sorely lacking.

After spending two hours on a question, we still cannot solve it and are always left with a $k$. Admittedly, I don't fully know what I am doing.

Evaluate $\displaystyle \sum_{k=1}^\infty\frac{x^k}{k(k+1)}$.

We do know that the answer is $1 + \dfrac{1-x}{x}\ln (1-x)$.

Any hints or help in the direction to go would be very much appreciated.

metamorphy
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Quinn
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1 Answers1

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Notice that $$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$$ and recall that for $|x|<1$, $$-\ln(1-x)=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}=\sum_{k=1}^{\infty}\frac{x^k}{k}$$ (which can be obtained by term-wise integration of $1/(1-x)=\sum_{k=0}^{\infty}x^k$). Therefore $$\begin{align}\sum_{k=1}^\infty\frac{x^k}{k(k+1)}&= \sum_{k=1}^\infty \frac{x^k}{k}-\sum_{k=1}^\infty\frac{x^{k}}{k+1}\\ &= \sum_{k=1}^\infty \frac{x^k}{k}-\sum_{k=0}^\infty\frac{x^{k}}{k+1}+1\\ &=\sum_{k=1}^\infty \frac{x^k}{k}-\frac{1}{x}\sum_{k=0}^\infty\frac{x^{k+1}}{k+1}+1\\ &=-\ln(1-x)-\frac{-\ln(1-x)}{x}+1\\ &=1 + \dfrac{1-x}{x}\ln (1-x).\end{align}$$

Robert Z
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