Geometric Series: Summing the Powers of x

[kezman]

Find the sum of the series:
$$\sum\limits_{n = 1}^\infty {nx^n }$$ if $$\left| x \right| < 1<br />$$


I thought maybe with the geometric form, but I am not sure.

 

[Geekster]

Find the sum of the series:
$$\sum\limits_{n = 1}^\infty {nx^n }$$

if
$$\left| x \right| < 1$$I thought maybe with the geometric form, but I am not sure.

Is it asking for a number, or just if the series converges?

 

[kezman]

I think for a general solution. It should converge.

 

[verd]

Is it asking you to find a power series representation??

 

[d_leet]

Is it asking you to find a power series representation??

It already is a power series... It's asking for an expression for the sum.

 

[Geekster]

You might start by writing out partial sums and see if that gets you anywhere...

 

[shmoe]

Find the sum of the series:
$$\sum\limits_{n = 1}^\infty {nx^n }$$ if $$\left| x \right| < 1<br />$$


I thought maybe with the geometric form, but I am not sure.

How does this differ from your usual geometric series?

 

[Curious3141]

Find the sum of the series:
$$\sum\limits_{n = 1}^\infty {nx^n }$$ if $$\left| x \right| < 1<br />$$


I thought maybe with the geometric form, but I am not sure.

Hint : Call the original series S. Write out the first five or so terms in the series. Divide the series by x to get a new series (S/x). Now take the difference between this new series and the original series (S/x - S), term by term and see what you end up with.

The other way to do it is to differentiate a geometric series, but that's more complicated and unnecessary.

 

[Pyrrhus]

Check https://www.physicsforums.com/showthread.php?t=73843"

 

[Curious3141]

Check https://www.physicsforums.com/showthread.php?t=73843"

Comparing series like these to derivatives of geometric series is a nice and interesting approach (I used to do this), but in most cases I've found that simply dividing or multiplying by x is an easier approach.

 

[shmoe]

May as well have a third approach:

$$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}\sum_{i=1}^{n}x^n$$

Change the order of summation (absolutely convergent series) then apply geometric series a couple of times. This is maybe the most complicated of the three, practice in rearranging summations never hurt.

 

[kezman]

thanks for all the hints.

The method I had to use is the derivative of the geometric series (similar to the one used for the maclaurin problem) using

$$<br /> \left( {\frac{1}{{1 - x}}} \right)^\prime = \sum\limits_{n = 0}^\infty {nx^{n - 1} } <br /> <br />$$