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We have the series $\sum_{n=1}^\infty nx^n$. The ratio test tells us simply the series is convergent when $|x|<1$ and divergent when $|x|>1$. Of course, when $|x|=1$, the series is $\sum n$ or $\sum -n$, which are obviously divergent. Thus, this series is convergent if and only if $-1<x<1$.

The difficulty comes when we want to find another expression of the function to which this series converges. When I first tried this exercise, I integrated the series, obtaining an antiderivative with no infinite sums, and then differentiated the result: $$\int\sum nx^n dx=\int x\sum nx^{n-1} dx$$ $$u=x\implies du=dx$$ $$dv=\sum nx^{n-1} dx \implies v=\sum\int nx^{n-1} dx=\sum x^n=\frac{x}{1-x}$$ $$\int\sum nx^n=\int x\sum nx^{n-1}=\frac{x^2}{1-x} -\int\frac{x}{1-x} dx=...$$

The rest are basic calculations. However, I've been told there's are more direct approach to this exercise, finding an easy expression of the partial sums. I hope you can help me find this other way.

Seven
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3 Answers3

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Let $S_n=\sum_{k=1}^{n} kx^{k}$. Then $xs_n=\sum_{k=1}^{n} kx^{k+1}=\sum_{k=2}^{n+1} (k-1)x^{k}=\sum_{k=2}^{n+1} kx^{k}-\sum_{k=2}^{n+1} x^{k}$. Hence $xS_n-S_n=(n+1)x^{n+1}-x-\sum_{k=2}^{n+1} x^{k}$. Write the geometric sum in the last term to get the value of $(x-1)S_n$. Divide by $x-1$ and let $n \to \infty$.

Kavi Rama Murthy
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Let: $$f\left(x\right)=1+x+x^{2}+x^{3}+\cdots=\frac{1}{1-x}$$ so that: $$f'\left(x\right)=\frac{1}{\left(1-x\right)^{2}}$$

Then apply: $$xf'\left(x\right)=x\left(1+2x+3x^{2}+\cdots\right)=\sum_{n=1}^{\infty}nx^{n}$$ to achieve that: $$\sum_{n=1}^{\infty}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$$

drhab
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  • It would be very nice to add some conditions to the formal operations in your post. – Marian G. Sep 05 '18 at 08:42
  • @MarianG. The problem is that I am not well-informed about that (anymore). It has been several decades ago that analysis was on my plate. With this answer I only show a route that can be taken, and I just hope that the OP can find his/her way with it. – drhab Sep 05 '18 at 08:55
  • Thank your for explanation. I understand. The post of Kavi Rama Murthy shows a more elemental way without any differentiation. – Marian G. Sep 05 '18 at 09:27
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$$x(x+2x^2+3x^3+4x^4+\cdots)=x^2+2x^3+3x^4+4x^5+\cdots \\=x+2x^2+3x^3+4x^4+5x^5+\cdots-(x+x^2+x^3+x^4+x^5+\cdots)$$

and

$$xS=S-\frac x{1-x}.$$