# Derive the sum of $\sum_{i=1}^n ix^{i-1}$

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.

• 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. – Peter Foreman Apr 15 at 12:40
• Minor notional detail: derivatives are written $\frac{d}{dx}$ not $\frac{d}{d}$. – DMcMor Apr 15 at 12:46
• DMcMor, fixed, thank you – Alex Apr 15 at 13:09
• The summation in the title is incorrect, the index is $i$, not $n$. – Yves Daoust Apr 15 at 13:13
• fixed as well... – Alex Apr 15 at 13:15

$$p_n(x):=\sum_{i=1}^n x^i$$ 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

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

• 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... – Alex Apr 15 at 13:27
• @Alex: do you see that in my answer ? – Yves Daoust Apr 15 at 13:28
• thank you, I really misunderstood your answer :) it is correct! – Alex Apr 15 at 13:35

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.

• Thank you for your reply! I probably misunderstand what does "closed form" mean. I've tried your answer, but this is incorrect as well... – Alex Apr 15 at 13:07
• 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. – Cameron Buie Apr 15 at 17:14
• 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. – Cameron Buie Apr 15 at 17:18