# Find out if the following series converges or diverges Sum(upper^infinity,lower n=1) sqrt(n) multiply by e^(-n)

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### 1 Answer

Here is one way to do it:

First consider the series

`sum_(n=1)^oo n e^(-n)`

We can use the integral test on this series:

Let `f(x)=xe^(-x)`

Then `f'(x)=1 e^(-x)+xe^(-x)*-1 = e^(-x)(x-x)`

Thus, if `x>1` then `f'(x)` is negative, so `f(x)` is decreasing

Also, if `x>0`, then `0<f(x)`

Thus we may use the integral test, and `sum n e^(-n)` converges if `int xe^(-x)dx` converges. So:

`int _1^oo xe^(-x)dx` = [-xe^(-x)-e^(-x)]_1^oo`

To see this, do integration by parts, with `u=x`, `dv = e^(-x)dx`

Certainly we may plug `x=1` into `-xe^(-x)-e^(-x)` with no problem

What we need to show is that:

`lim_(x->oo) -xe^(-x)-e^(-x)` exists and is finite.

As `x->oo` `e^(-x) = (1)/(e^x) -> 0 ` For the other piece, we need to use l'Hopitals:

`lim_(x->oo) -xe^(-x) = lim_(x->oo) (-x)/(e^x) = lim_(x->oo) (-1)/(e^x) = 0`

Thus, all the pieces converge, and the integral converges. So the series converges as well.

Now, we have:

`sqrt(n) <= n` (if `n>=1`), so:

`0 < sqrt(n)e^(-n) <= n e^(-n)`

Since `sum n e^(-n)` converges, by comparison test, `sum sqrt(n)e^(-n)` must converge as well.