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The series $$\sum^\infty_{k=1}kx^k$$converges to $$\frac{z}{(1-z)^2}$$ Could anyone give me a hint on how this convergence can be proven? I am especially puzzled as the $k$ on the left seems to disappear completely in the process of convergence.

Alessandro
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Jonas
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  • The $(1-z)^2$ that you see in the right, take it multiplying to the left and expand. All terms cancel except $z$. – plop May 23 '21 at 20:33
  • Let $S_n=\sum^n_{k=1}kx^k$ and look at $$S_{n+1}-xS_n$$then take $n\rightarrow\infty$ (under some restriction on the value of $x$, so the limit is finite) – Alessandro May 23 '21 at 20:36

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Hint: $$\sum_{k=1}^\infty kz^k=\sum_{k=1}^\infty(k+1)z^k-z^k=\sum_{k=1}^\infty(k+1)z^k-\sum_{k=1}^\infty z^k=\frac{d}{dz}\left[\sum_{k=1}^\infty z^{k+1} \right]-\sum_{k=1}^\infty z^k$$

Even easier: $$\sum_{k=1}^\infty kz^k=z\sum_{k=1}^\infty kz^{k-1}=z\frac{d}{dz}\left[\sum_{k=1}^\infty z^k \right]$$

Rhys Hughes
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    Thank you! This is helpful. But I have 2 questions: is this true for any z? And I still don't quite see how I get from here to the convergence. – Jonas May 23 '21 at 21:28
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    @Jonas Only true for $\lvert z\rvert <1$. – vitamin d May 23 '21 at 21:40
  • We have $$\sum_{k=0}^\infty z^k=\lim_{n\to\infty}\frac{1-z^{n+1}}{1-z}$$ which evaluates to $\frac{1}{1-z}$ when $|z|<1\implies z^n\to 0$ (as vitamin d notes). Both of the sums at the end can be put in terms of the geometric series above. – Rhys Hughes May 23 '21 at 21:50
  • @Jonas $$\sum_{k=1}^\infty z^k=\sum_{k=0}^\infty z^{k+1}=z\sum_{k=0}^\infty z^k=?$$ – Rhys Hughes May 23 '21 at 21:52
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    This clearly evaluates to z/(1-z). – Jonas May 23 '21 at 22:02
  • Exactly. Can you do the other one? – Rhys Hughes May 23 '21 at 22:02
  • If I try to evaluate the above-mentioned differential, I have a problem, cause in order to bring it to the form needed (starting with k=0) I would have to divide the term by z^2 and then the differentiation is no longer straightforward, as I would have to use the quotient rule. – Jonas May 23 '21 at 22:09
  • You should have the sum evaluate to $\frac{z^2}{1-z}$ and then use the quotient rule to differentiate it. – Rhys Hughes May 23 '21 at 22:17
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    Thank you! Now I received the desired result. And thank you also for being so patient and explaining the details! – Jonas May 23 '21 at 22:25