How to calculate sum of power series? [duplicate]

I'm trying to work out sum of this series $$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots$$

I know one method is to do substitutions and getting the series into a form of a known series. So far I've converted the series into $$1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \ldots$$ where $x=2$ and I'm trying to get it into the form of the $\ln(1+x)$ series somehow. I have tried differentiating, integrating and nothing is working out. The closest I got is by inverting which gave me $1 + \frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} \ldots$

Now I'm just lost and have no idea what to do.

The other idea I had was converting it into $$\Large{\sum_{n=1}^\infty{\frac{n}{2^{n-1}}}},$$ but I have no idea how to do anything further to it. How would you do this? Thnx.

marked as duplicate by David K, colormegone, JMoravitz, Travis, user296602 Jan 5 '16 at 1:43

• You could try integrating? If it were in the form $x+2x^2+3x^3+4x^4\cdots$, we could integrate, find the solution, then take the derivative of that. – Simply Beautiful Art Jan 5 '16 at 0:58

Cherry pick the terms into series you can sum. THus you know how to do $$1 + \frac{1}{2} + \frac{1}4 + \frac{1}{8}+ \frac{1}{16} \cdots = 2$$ and you are left with $$\frac{1}{2} + \frac{2}4 + \frac{3}{8} + \frac{3}{16} \cdots$$ so cherry pick again: $$\frac{1}{2} + \frac{1}4 + \frac{1}{8}+ \frac{1}{16} \cdots = 1$$ leaving $$\frac{1}4 + \frac{2}{8} + \frac{3}{16} \cdots$$ And next you will get $\frac{1}{2}$ and so forth. Then sum all of those sums, to get the answer, which is somewhere around $4$.

• "Somewhere around"? :-) – Brian Tung Jan 5 '16 at 0:43

You started with this, which I will set equal to $k$:

$$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots=k$$

Use the substitution $x=\frac12$.

$$1+2x+3x^2+4x^3+\ldots=k$$

This is where it gets a little bit confusing. We have to integrate the left side, find a solution to it, then differentiate the solution.

$$\int 1+2x+3x^2+4x^3+\ldots dx=x+x^2+x^3+x^4\ldots=\sum_{n=1}^{\infty}x^n$$

We have a well known solution for this:

$$\sum_{n=1}^{\infty}x^n=\frac1{1-x}$$

Now return to what we wanted in the first place:$$\int1+2x+3x^2+4x^3\ldots dx=\frac1{1-x}$$

Differentiate both sides to get:

$$1+2x+3x^2+4x^3\ldots=\frac1{(1-x)^2}$$

Use $x=\frac12$.

$$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots=\frac1{(1-\frac12)^2}=4=k$$

Note that this is probably going to arouse some conversation, so I will try my best to explain why I can do this.

If I differentiate and integrate two functions $f(x)$ and $g(x)$ that are equal to each other, then we will have $f'(x)=g'(x)$ if and only if $f(x)+C=g(x)$ for all $x$.

$$f(x)+C=g(x)$$

$$f'(x)=g'(x)$$

You can support this with a bit of graphing.

You may also note that I ignored the constant of integration when I integrated, only because I would differentiate in the near future, which would negate my constant of integration.

This method may produce what you might consider the correct answer, and numerical checks say I am correct.