Infinite series problem

From: Matt Gutting (tchrmatt_at_yahoo.com)
Date: 03/14/05 

Date: Sun, 13 Mar 2005 22:28:28 -0500

This was posted yesterday on a web site I frequent; it was a homework problem
assigned to one of the posters. I attempted to give him some help on one of
the infinite series, but I'd like to make sure what I told him was right, as
it looks a bit odd to me.

The problem was:

Calculate Sum(i=1 to infinity) {-1^i)(i)(i+1)/(2^i)

(I hope that's written clearly).

I decided to multiply out the i terms and get a sum in i^2 and another in i.
I was stumped for a while over the term in i (sum of i/(2^i)) then realized
you could do

-1/2 = -1/2

  2/4 = 1/4 + 1/4

-3/8 = -1/8 + -1/8 + -1/8

etc. - coming up to sum(-1/2)^i + -1/2(sum(-1/2)^i) + 1/4....

that is, sum(i=1 to infinity)((-1/2)^i) * sum(i=0 to infinity)((-1/2)^i)

The first term appears to sum to -1/3, the second to 2/3, and I got the product
(therefore the sum of the first series) as -2/9. Offhand this *looks*
reasonable, but I'd like to figure out an independent method to check it.
If it works I can (fairly) easily apply the same technique to the i^2 sum
(I just have to remember to multiply successive series by successive odd
numbers).

Again, am I on the right track here? And is there another way of arriving at
the same sum? (I'm assuming that the sum does converge, since the numerator
is a polynomial and the denominator is an exponential, but I haven't
actually *checked* that yet.)

Matt

Relevant Pages

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