What is the general form and convergence of (k+1)r^k?

[Caspian]

[solved] Sum of k x^k?

I happened upon a thread in a math forum, where someone asserted that this is true:

\sum_{k=0}^\infty (k+1) \left(\frac{5}{6}\right)^k = 36

I suppose this makes intuitive sense. But if it's true, it must have a general form. I.e.,

\sum_{k=0}^\infty (k+1) r^k = ?

Now, I know that the geometric series converges like so:

\sum_{k=0}^\infty r^k = \frac{1}{1-r}

But by multiplying by (k+1) inside the summation completely changes things. Is there a name for this series? Is it true that it converges? If so, what does it converge to?

This question won't stop plaguing me. Since I don't know what this series is called, I'm having a hard time searching for it on the Internet.

Thanks!

 

[jgens]

Well, I won't derive the general case for you but I will show you how to evaluate this particular sum (although, I'm not sure that I agree with the answer that someone else provided) . . .

S = \sum_{k=1}^{\infty}\frac{k5^k}{6^k} = \frac{5}{6} + 2\left(\frac{5^2}{6^2}\right) + \dots = \frac{5}{6} \left [ 1 + 2\left(\frac{5}{6}\right) + 3\left(\frac{5^2}{6^2}\right) + \dots \right ]

Therefore, we know that

S = \frac{5}{6} \left [ 1 + 2\left(\frac{5}{6}\right) + 3\left(\frac{5^2}{6^2}\right) + \dots \right ] = \frac{5}{6} \left ( 1 + \sum_{k=1}^{\infty}\frac{(k + 1)5^k}{6^k} \right )

With some simple manipulations, we can put this last sum into a more desirable form

\sum_{k=1}^{\infty}\frac{(k + 1)5^k}{6^k} = \sum_{k=1}^{\infty}\frac{k5^k}{6^k} + \sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k = S + \sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k

Now, using this expression we find that

S = \frac{5}{6} + \frac{5S}{6} + \frac{5}{6}\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k

From which it follows that

S = 5 + 5\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k

Using the formula for the sum of a geometric series, we can evaulate the last sum and find the value of S

S = 5 + 5\left(\frac{1}{1 - \frac{5}{6}}\right) = 5 + 5(6) = 35

Edit: Aside from any significant mistakes I may have made in evaluating that particular sum, you should also note that I evaluated the sum using very informal methods.

 

[elibj123]

Convergence issues aside (actually it converges for every complex r with absolute value less than 1, and converges uniformly on every disk with a radius less than 1)
A good insight is that it is a derivative of a more simple series:

\frac{d}{dr}\sum^{\infty}_{k=0}r^{k+1}=\sum^{\infty}_{k=0}(k+1)r^{k}

And you know that:

\sum^{\infty}_{k=0}r^{k+1}=r\sum^{\infty}_{k=0}r^{k}=\frac{r}{1-r} (For any r for which it converges)

And therefore your sum is given by

S(r)=\frac{d}{dr}(\frac{r}{1-r})=\frac{(1-r)+r}{(1-r)^{2}}=\frac{1}{(1-r)^{2}}

Which also fits your special case r=5/6.

 

[jgens]

Wow, that's much simpler. It really makes me wish that I was more proficient with infinite series.

 

[Caspian]

That was just what I was looking for -- not only the solution, but a tool to use in the future :).

Thank you so much!