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How can I prove for $|x|<1$ the following equality?

$$\frac{1}{(1-x)^2} = \sum_{k=1}^\infty kx^{k-1}$$

I tried to write

$\sum_{k=1}^\infty kx^{k-1} = (\sum_{k=0}^\infty k+1) (\sum_{k=0}^\infty x^{k})$ = $(\sum_{k=0}^\infty k+1)(\frac{1}{1-x}) $

But the row $\sum_{k=0}^\infty k+1$ diverges. Can someone help me? Thx

Jack D'Aurizio
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Marc
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  • There are multiple ways to prove the equality. I just saw a very similar question. Have a look at https://math.stackexchange.com/questions/2542916/let-gx-frac11-x2-prove-that-gx-sum-n-0-inftyn1xn – Math Lover Nov 29 '17 at 15:48
  • $$\sum a_n b_n \neq \left(\sum a_n\right)\left(\sum b_n\right)$$ – Jack D'Aurizio Nov 29 '17 at 15:48
  • But $k x^{k-1}$ is the derivative of $x^k$, so your power series is the derivative of the geometric series $1+x+x^2+x^3+\ldots=\frac{1}{1-x}$. – Jack D'Aurizio Nov 29 '17 at 15:49
  • See also: https://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot – lab bhattacharjee Nov 29 '17 at 16:04

2 Answers2

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The way you tried to solve it is not mathematical correct.

$$ \sum_{k=1}^{\infty}kx^{k-1} = \frac{d}{dx}\sum_{k=1}^{\infty}x^{k} =\frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2} $$ Using the principle of a geometric series.

As pointed out by @Botond, technically the infinite sum should start from $k=0$ but is a constant with respect to $x$ so is ignored as follows $$ \sum_{k=0}^\infty \frac{d}{dx} x^k = \frac{d}{dx} x^0 + \sum_{k=1}^\infty \frac{d}{dx} x^k = 0 +\sum_{k=1}^\infty \frac{d}{dx} x^k $$

Chinny84
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Differentiation theorem for power series : A power series can be differentiated term-by-term within the interval of convergence. In fact, if $$f(x)=\sum_{n=0}^{\infty} a_n x^n, \; \text {then} \; f'(x)=\sum_{n=1}^{\infty} na_n x^{n-1} \; \text {for} \; |x| \lt R.$$ Both series have same radius of convergence $R$. (From Introduction to Real Analysis by Bartle and Sherbert.)

Can you take it from here? You have to find out $f(x)$ whose derivative is $\frac {1}{1-x^2}$ and which is sum of a well-known geometric series on $|x| \lt 1$.

Error 404
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