0
$\begingroup$

I am trying to find the convergent value for the following summation:

$$\sum_{a=0}^∞ a \left( \frac{x-1}{x}\right)^{\!a}$$

This sum seems to be convergent by ratio test as $$\frac{x-1}{x} < 1$$ but I am unsure of how to deal with the auxiliary $a$ term being multiplied in the summation. I am aware of the geometric series formula for infinite sums in the general case, but since we have a varying constant inside the summation, the series summation does not seem to apply. Could someone elaborate the steps to take to solve this summation?

Thanks in advance -- all help is greatly appreciated!

$\endgroup$
2

1 Answer 1

1
$\begingroup$

Let $y= \frac{x-1}{x}$ $$\sum_{a=0}^\infty a \left( \frac{x-1}{x}\right)^{a}=\sum_{a=0}^\infty a y^{a}=y\sum_{a=0}^\infty a y^{a-1}=y\frac d {dy}\left(\sum_{a=0}^\infty y^{a} \right)$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .