Evaluate a sum

[intervade]

Alright, so I'm given the following sum:

\(\displaystyle \sum_{n=0}^{\infty }(n+1)(\frac{1}{2})^n\)

Now, I need to find its value. I was told to start with a series that I already know so I'm sort of using a brute force approach. I have a list of power series for some basic functions but I don't really know where to start to be honest. I know the the function sums up to 4.

I also know that (using a calculator):

\(\displaystyle \sum_{n=0}^{\infty }(\frac{1}{2})^n = 2\) and
\(\displaystyle \sum_{n=0}^{\infty }n(\frac{1}{2})^n = 2\)

( Clearly 2 + 2 = 4, but I would like to show why )

But I don't know if thats where I should be headed. Any suggestions?

 

[daon]

Here's a little "naive calculus":

\(\displaystyle \frac{d}{dx} \sum_{n=0}^{\infty} x^{n} = \sum_{n=0}^{\infty} nx^{n-1}\)

That LHS is also equal to:

\(\displaystyle \frac{d}{dx}\frac{1}{1-x} = \frac{1}{(1-x)^2}\)

 

[intervade]

Hmm I dont still can't quite follow how that helps me. Could you elaborate a little more?

 

[daon]

Try plugging in 1/2 for x yet? Playing with the indicies a bit you see:

\(\displaystyle \sum_{n=0}^{\infty}(n+1)x^n = \sum_{n=0}^{\infty}(n)x^{n-1} = \frac{1}{(1-x)^2}\)

 

[BigGlenntheHeavy]

\(\displaystyle All \ in \ all, \ a \ neat \ trick.\)

\(\displaystyle As \ they \ say \ (whoever \ they \ are), \ it's \ all \ in \ the \ manipulation.\)

 

[Deleted member 4993]

You know why:

\(\displaystyle \sum_{n=0}^{\infty}\left ( \frac{1}{2}\right)^n = 2\)

correct??

Spoiler: :2o8u01xo

- it is a geometric series of ratio = 1/2 and first term = 1[/spoiler:2o8u01xo]

 

[galactus]

Here is a tutorial on what the Big G is talking about.

Say we want to find \(\displaystyle \sum_{n=0}^{\infty}\frac{n^{2}}{3^{n}}\)

Start with the formula for a geometric series: \(\displaystyle \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}\)

\(\displaystyle \frac{d^{2}}{dx^{2}}\left[\frac{1}{1-x}\right]=\sum_{n=2}^{\infty}n(n-1)x^{n-2}\)

\(\displaystyle x^{2}\cdot \frac{d^{2}}{dx^{2}}\left[\frac{1}{1-x}\right]=\sum_{n=0}^{\infty}n(n-1)x^{n}\)

So, we have:

\(\displaystyle (n^{2}-n)x^{n}=n^{2}x^{n}-nx^{n}\)

\(\displaystyle x\cdot \frac{d}{dx}\left[\frac{1}{1-x}\right]=\sum_{n=0}^{\infty}nx^{n}\)

Which gives us \(\displaystyle \sum_{n=0}^{\infty}n^{2}x^{n}=\frac{-x(x+1)}{(x-1)^{3}}\)

\(\displaystyle \sum_{n=0}^{\infty}\frac{n^{2}}{3^{n}}=\frac{-\frac{1}{3}(\frac{1}{3}+1)}{(\frac{1}{3}-1)^{3}}\)

Therefore, \(\displaystyle \sum_{n=0}^{\infty}\frac{n^{2}}{3^{n}}=\boxed{\frac{3}{2}}\)

Does that help illustrate?. We can use differentiating (and integrating) of the geometric series to do all sorts of things.

 

[soroban]

Hello, intervade!

\(\displaystyle \text{Evaluate: }\;\sum_{n=0}^{\infty } \frac{n+1}{2^n}\)

\(\displaystyle \begin{array}{ccccc}\text{We are given:} & S &=& \dfrac{1}{1} + \dfrac{2}{2} + \dfrac{3}{2^2} + \dfrac{4}{2^3} + \dfrac{5}{2^4} + \dfrac{6}{2^5} + \hdots \\ \\[-2mm] \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \qquad \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \dfrac{5}{2^5} + \hdots \\ \end{array}\)

. . \(\displaystyle \begin{array}{cccc}\text{Subtract:} &\quad \frac{1}{2}S &=& \;\underbrace{1 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + \dfrac{1}{2^5} + \hdots}_{\text{geometric series}} \end{array}\)

. . \(\displaystyle \text{The geometric series has sum: }\:\frac{1}{1-\frac{1}{2}} \:=\:2\)


\(\displaystyle \text{Therefore: }\;\tfrac{1}{2}S \:=\:2 \quad\Rightarrow\quad \boxed{S \:=\:4}\)