# Thread: Sums using the integral test

1. ## Sums using the integral test

sum e^n/(1+e^2n)
n = 1 to infinity

By using the integral test I need to integrate this to determine whether it will converge or diverge.

2. Originally Posted by ur5pointos2slo
sum e^n/(1+e^2n)
n = 1 to infinity

By using the integral test I need to integrate this to determine whether it will converge or diverge.

$\int \frac{e^n}{1+e^{2n}} dn$

use substution,

let $u = e^{n}$, then $du = e^n dn \implies dn = \frac{du}{e^n}$

Do the rest! show where you are getting stuck

3. Originally Posted by harish21
$\int \frac{e^n}{1+e^{2n}} dn$

use substution,

let $u = e^{n}$, then $du = e^n dn \implies dn = \frac{du}{e^n}$

Do the rest! show where you are getting stuck
My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?

4. Originally Posted by ur5pointos2slo
My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?
scratch that stupid me. Ok we are using a u-substitution. I am confused on where to go next. There must be a trick using exponential rules?

5. Originally Posted by ur5pointos2slo
My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?
$du = e^n dn \implies dn = \frac{du}{e^n}$

$\therefore \int \frac{e^n}{1+e^{2n}} dn$

$= \int \frac{1}{1+u^2} du = tan^{-1}(u)$

plug back $u = e^n$, you have:

$[tan^{-1}(e^n)]_1^\infty$

now find out if the series converges or not!

6. Originally Posted by harish21
$du = e^n dn \implies dn = \frac{du}{e^n}$

$\therefore \int \frac{e^n}{1+e^{2n}} dn$

$= \int \frac{1}{1+u^2} du = tan^{-1}(u)$

plug back $u = e^n$, you have:

$[tan^{-1}(e^n)]_1^\infty$

now find out if the series converges or not!
That seemed too easy. I couldn't see it until you pointed it out. So it equals Pi/4 and it converges.

7. Originally Posted by ur5pointos2slo
That seemed too easy. I couldn't see it until you pointed it out. So it equals Pi/4 and it converges.
$\frac{\pi}{4}$??

$[tan^{-1}(e^n)]_1^\infty = (tan^{-1}(e^{\infty})) - (tan^{-1}(e^1)) = (tan^{-1}(\infty)) - (tan^{-1}(e)) = \frac{\pi}{2} - tan^{-1}(e)$, This is not equal to $\frac{\pi}{4}$

8. Originally Posted by harish21
$\frac{\pi}{4}$??

$[tan^{-1}(e^n)]_1^\infty = (tan^{-1}(e^{\infty})) - (tan^{-1}(e^1)) = (tan^{-1}(\infty)) - (tan^{-1}(e)) = \frac{\pi}{2} - tan^{-1}(e)$
oops forgot the exponential...so Pi/2 - tan^-1(e) converges.

9. Originally Posted by ur5pointos2slo
oops forgot the exponential...so Pi/2 - tan^-1(e) converges.
Yes!