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

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The discussion centers on the convergence and general form of the series ∑(k+1)r^k. A participant confirms that the series converges for |r| < 1 and derives the formula S(r) = 1/(1-r)², which applies to the specific case where r = 5/6, yielding a sum of 36. The conversation highlights the relationship between this series and the geometric series, emphasizing that it can be evaluated using derivatives. Participants express satisfaction with the clarity and utility of the derived formula for future calculations. The thread concludes with an appreciation for the insights gained regarding infinite series.
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[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!
 
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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.
 
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.
 
Wow, that's much simpler. It really makes me wish that I was more proficient with infinite series.
 
That was just what I was looking for -- not only the solution, but a tool to use in the future :).

Thank you so much!
 
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