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Evaluate the sum: $$\sum^{\infty}_{k=0}(k+1)x^{k}$$

Now my attempt was to start with a series we know, like the geometric series. so $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$$ I also think there is a generic form of the form $$\sum_{k=0}^{\infty}a_kx^k=\frac{a}{1-x}$$

I am not exactly sure how to manipulate this known series to reach some conclusive summation for the sum I am being asked to evaluate. I have done simpler examples where you simply differentiate and than multiply through by x and the solution is there. But I do not know what to do here to find the sum, I would appreciate some tips or advice.

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$$\begin{align} f(x)&=\sum_{k=0}^\infty x^{k+1}&&=\frac{x}{1-x}\\ f'(x)&=\sum_{k=0}^\infty (k+1)x^k&&=\left(\frac{x}{1-x}\right)'\\ \end{align}$$

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  • $\begingroup$ okay so I differentiate and I get $$\frac{1}{(1-x)^2}$$ now what value do I plug in in order to find the actual sum of this series or is the sum just $$\frac{1}{(1-x)^2}$$. I was under the impression that it is suppose to be a number. $\endgroup$ – B ry Oct 27 '16 at 18:20
  • $\begingroup$ @Bry Yeah that's the sum. It's is dependent on $x$ so it can't be a single number. If you had $x=0.5$, you'd get a sum of $4$ but with $x=0.75$, you get a sum of $16$. That's partly why I wrote it as $f(x)$, since I think it's a bit clearer that way. But as MitchellSpector says, you can do it without needing any calculus. Cool? $\endgroup$ – Jam Oct 27 '16 at 18:24
  • $\begingroup$ what about something like $$\sum_{k=0}^{\infty}(-1)^k(k+1)x^{2k}$$ $\endgroup$ – B ry Oct 27 '16 at 18:37
  • $\begingroup$ @Bry Not sure about that one off the top of my head but try reading the link SeanRobertson posted about Generating Functions, which tells you a lot about these types of series. $\endgroup$ – Jam Oct 27 '16 at 18:46
  • $\begingroup$ @Bry I added an addendum to my answer to show how $\sum_{k=0}^{\infty}(-1)^k(k+1)x^{2k}$ can be computed from the first sum. $\endgroup$ – Mitchell Spector Oct 27 '16 at 19:54
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There's no need to use derivatives.

Set $$S=1+2x+3x^2+4x^3+\dots$$ and subtract

$$xS=\hphantom{1+2}x+2x^2+3x^3+\dots$$ to get

$$(1-x)S=1+x+x^2+x^3+\dots = \frac1{1-x},$$

so your sum is

$$S=\frac1{(1-x)^2}.$$

This is valid for $-1\lt x\lt 1.$


Addendum:

As for $\sum_{k=0}^{\infty}(-1)^k(k+1)x^{2k},$ which you asked about in a comment, this follows from the first sum:

\begin{align}\sum_{k=0}^{\infty}(-1)^k(k+1)x^{2k} &= \sum_{k=0}^{\infty}(k+1)(-x^2)^{k} \\&=\frac1{(1-(-x^2))^2} \\&==\frac1{(1+x^2)^2}, \end{align}

again for $-1\lt x\lt 1.$

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To avoid repetition, we know that

$$f(x)=\sum_{k=0}^\infty(k+1)x^k=\frac1{(1-x)^2}$$

As for the last part, if we have

$$f(x)=\sum_{k=0}^\infty a_kx^k$$

then $f(x)$ is analytic, assuming it converges for some $x$ with radius of convergence greater than $0$.

According to Taylor's theorem, we then have

$$f(0)=a_0\\f'(0)=a_1\\f''(0)=2a_2\\f'''(0)=6a_3$$

More generally, we have

$$f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k$$

where $f^{(n)}(x)$ is the $n^{\text{th}}$ derivative of $f$.

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