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For the series

$$1 + 2x + 3x^2 + 4x^3 + 5x^4 + ... + nx^{n-1}+... $$

and $x \ne 1, |x| < 1$.

I need to find partial sums and finally, the sum $S_n$ of series. Here is what I've tried:

  1. We can take a series $S_2 = 1 + x + x^2 + x^3 + x^4 + ...$ so that $\frac{d(S_2)}{dx} = S_1$ (source series).
  2. For the $|x| < 1$ the sum of $S_2$ (here is geometric progression): $\frac{1-x^n}{1-x} = \frac{1}{1-x}$
  3. $S_1 = \frac{d(S_2)}{dx} = \frac{d(\frac{1}{1-x})}{dx} = \frac{1}{(1-x)^2}$

But this answer is incorrect. Where is my mistake? Thank you.

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    $\begingroup$ This is correct for the sum to $\infty$ but you need to take the derivative of $\frac{1-x^n}{1-x}$ for the partial sum. $\endgroup$ – Peter Foreman Apr 15 at 12:40
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    $\begingroup$ Minor notional detail: derivatives are written $\frac{d}{dx}$ not $\frac{d}{d}$. $\endgroup$ – DMcMor Apr 15 at 12:46
  • $\begingroup$ DMcMor, fixed, thank you $\endgroup$ – Alex Apr 15 at 13:09
  • $\begingroup$ The summation in the title is incorrect, the index is $i$, not $n$. $\endgroup$ – Yves Daoust Apr 15 at 13:13
  • $\begingroup$ fixed as well... $\endgroup$ – Alex Apr 15 at 13:15
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$$p_n(x):=\sum_{i=1}^n x^i$$ is a polynomial, which you can differentiate term-wise, giving the polynomial

$$p'_n(x):=\sum_{i=1}^n ix^{i-1}.$$

At the same time, $p(x)$ is the sum of terms of a geometric series, and for $x\ne1$,

$$p_n(x)=\frac{x^{n+1}-1}{x-1}-1.$$

Then, for all $x\ne1$,

$$p'_n(x)=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{(x-1)^2}.$$


The limit exists for all $|x|<1$, and

$$p'_\infty(x)=\dfrac1{(x-1)^2}.$$

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  • $\begingroup$ Thank you! I liked your way of solving, but as I've mentioned in my question, the answer $S_n = \frac{1}{(x-1)^2}$ is incorrect... $\endgroup$ – Alex Apr 15 at 13:27
  • $\begingroup$ @Alex: do you see that in my answer ? $\endgroup$ – Yves Daoust Apr 15 at 13:28
  • $\begingroup$ thank you, I really misunderstood your answer :) it is correct! $\endgroup$ – Alex Apr 15 at 13:35
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You've got a good notion!

Integrating the partial sum

$$1+2x+\cdots nx^{n-1}$$ gives you $$C+x+x^2+\cdots x^n,$$ for some constant $C,$ which is $$C-1+\frac{1-x^{n+1}}{1-x}.$$ Then, taking the derivative using the quotient rule gets you $$\begin{eqnarray}\frac{-(n+1)(1-x)x^n+1-x^{n+1}}{(1-x)^2} &=& \frac{-(n+1)x^n+(n+2)x^{n+1}+1-x^{n+1}}{(1-x)^2}\\ &=& \frac{nx^{n+1}-(n+1)x^n+1}{(1-x)^2}\end{eqnarray}$$ for your partial sum's closed form.

You've correctly found the closed form of the limit of the partial sums.

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    $\begingroup$ Thank you for your reply! I probably misunderstand what does "closed form" mean. I've tried your answer, but this is incorrect as well... $\endgroup$ – Alex Apr 15 at 13:07
  • $\begingroup$ A "closed form" of a sum is one that doesn't use the $\Sigma.$ For example, the closed form of $\sum_{k=0}^n x^k$ is $\frac{1-x^{n+1}}{1-x}$ so long as $x\ne 1.$ The closed form of $\sum_{k=0}^\infty x^k$ is $\frac{1}{1-x}$ whenever $|x|<1.$ The answer is correct. See here for verification. $\endgroup$ – Cameron Buie Apr 15 at 17:14
  • $\begingroup$ Ah! I missed that the indices were supposed to range from $0$ to $n-1,$ rather than from $0$ to $n.$ I will adjust my answer accordingly now. $\endgroup$ – Cameron Buie Apr 15 at 17:18

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