## 0568 #51

Forum for the GRE subject test in mathematics.
jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

### 0568 #51

#51 If Floor(x) denotes the greatest integer not exceeding x, then Integral(Floor(x)*Exp(-x))dx from 0 to infinity = ?

The answer turns out to be 1/(e-1).

I can get it to Sum(n*Exp(-n)*(1-1/e)) from n = 0 to infinity, which Mathematica confirms is 1/(e-1), I just have no idea to find the sum of that infinite series by hand. Off the top of my head, I can only think of the sum of a geometric series or p-series where p = 2.

Any help would be appreciated.

mag487
Posts: 13
Joined: Thu Aug 06, 2009 3:59 am

### Re: 0568 #51

jayre wrote:#51 If Floor(x) denotes the greatest integer not exceeding x, then Integral(Floor(x)*Exp(-x))dx from 0 to infinity = ?

The answer turns out to be 1/(e-1).

I can get it to Sum(n*Exp(-n)*(1-1/e)) from n = 0 to infinity, which Mathematica confirms is 1/(e-1), I just have no idea to find the sum of that infinite series by hand. Off the top of my head, I can only think of the sum of a geometric series or p-series where p = 2.

Any help would be appreciated.
Factor out the (1 - 1/e) from the series. You end up with r + 2r^2 + 3r^3 + 4r^4 + ... where r = e^-1. We can re-express this as r*[(1 + r^2 + r^3 + ...) + (r + r^2 + r^3 + ...) + (r^2 + r^3 + r^4) + ...] = r*[(1 + r^2 + r^3 + ...) + r(1 + r^2 + r^3 + ...) + r^2(1 + r + r^2 + ...) + ...] = r*[1/(1 - r) + r/(1 - r) + r^2/(1 - r) + ...] = r/(1 - r)^2. Multiplying this by (1 - 1/e) = (1 - r) gives the answer.

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

### Re: 0568 #51

So, in general, Sum(n*r^n) = r/(1 - r)^2 for |r|< 1 ?

Thanks for the help, btw!

mag487
Posts: 13
Joined: Thu Aug 06, 2009 3:59 am

That's right.