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In Calculus, how do I calculate this sum?

$$f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$$

This is what I did so far: $$ f(x) = 2x \cdot \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^{n-1}} $$

Therefore: $$ \frac{\int{f(x)}}{2x} = \sum_{n=0}^{\infty} {2^n \cdot x^n}$$

But I have no idea where to continue from here!

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    $\begingroup$ Write the terms of the sum as $n(2x)^n$ and see this similar question for ideas. $\endgroup$ – David Mitra Apr 30 '13 at 14:52
  • $\begingroup$ Thank you! I didn't even notice that haha! $\endgroup$ – TheNotMe Apr 30 '13 at 14:54
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    $\begingroup$ The part after therefore is wrong. If you integrate, you cannot pull $x$ from the RHS in front of the integral. If you first divide and then integrate, the LHS is $\int \frac{f(x)}{x}$... The basic idea is right, but you don't need and shouldn't bring $x$ in front of the sum, just integrate in the first term. $\endgroup$ – N. S. Apr 30 '13 at 16:41
  • $\begingroup$ @N.S. you are correct, thank you! $\endgroup$ – TheNotMe Apr 30 '13 at 21:31
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$$\sum_{n=1}^\infty nt^n=t\frac{d}{dt}\left(\sum_{n=1}^\infty t^n\right)=t\frac{d}{dt}\left(\frac{t}{1-t}\right)=\frac{t}{(1-t)^2}$$

Now with $t=2x$ we find $$\sum_{n=1}^\infty n(2x)^n=\frac{2x}{(1-2x)^2}$$

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  • $\begingroup$ Looks good to me! $\endgroup$ – Namaste Jul 20 '14 at 11:46
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I believe you can sum the series at the point you left off, $$ \frac{\int f(x) dx}{x} = 1 + 2x + (2x)^2 + (2x)^3 + ... = \frac{1}{1-2x} $$ So, we work back to $f$ by itself, $$ \int f(x) dx = \frac{x}{1-2x} $$ We take derivatives, $$ f(x) = \frac{(1-2x) - x (-2)}{(1-2x)^2} = \frac{1}{(1-2x)^2} $$

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  • $\begingroup$ Note that you start at a wrong spot, since the OP made a mistake. $\endgroup$ – N. S. Apr 30 '13 at 16:42
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Here is a more elementary proof, which doesn't use the differentiability/integrability property of power series:

Let

$$S_m= \sum_{n=0}^{m} {n \cdot 2^n \cdot x^n}$$

Then

$$2x S_m =\sum_{n=0}^{m} {n \cdot 2^{n+1} \cdot x^{n+1}}=\sum_{k=1}^{m+1} {(k-1) \cdot 2^{k} \cdot x^{k}}$$ $$=\left(\sum_{k=1}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(\sum_{k=1}^{m+1} 2^{k} \cdot x^{k}\right)=\left(\sum_{k=0}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$ $$2xS_m=\left(S_m+(m+1)2^{m+1}x^{m+1}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$

Solving for $S_m$ yields:

$$S_m=\left(2x\cdot\frac{1-(2x)^{m+1}}{(1-2x)^2}\right)-\frac{(m+1)(2x)^{m+1}}{1-2x}$$

now, $S_m$ is convergent if and only if $(2x)^{m+1} \to 0$ if and only if $|2x|<1$.

In this case

$$\lim S_m= 2x\cdot\frac{1}{(1-2x)^2}$$

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