# Integral Test of e^n/(e^(2n)+9)

1. May 20, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Using the integral test determine whether or not the following sum diverges. $$\sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9}$$

2. Relevant equations
3. The attempt at a solution
$$\sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9}=\sum_{n=1}^{\infty} \frac{e^n}{({e^{n})}^2+9} \\ \int^{\infty} \frac{e^n}{({e^{n})}^2+9} \cdot dn; \hspace{0.5cm} Let \space u=e^n; \hspace{0.5cm} du=e^n\cdot dn \\ \int^{\infty} \frac{du}{u^2+9}; \hspace{0.5cm} Let \space u=3\tan\theta \hspace{0.5cm} du=3\sec^2\theta \\ \int^{(\frac{\pi}{2})^-} \frac{3\sec^2\theta}{9\tan^2\theta+9}=\frac{1}{3} \int^{(\frac{\pi}{2})^-} 1\cdot d\theta=\frac{\pi}{6}$$ Since I didn't get infinity in my answer this sum converges however my answer key says that it does diverge, did I miss something?

2. May 20, 2015

### SammyS

Staff Emeritus
What is the lower limit for the integral?

It is true that this series converges.

You have left dθ out of a couple of expressions.

Last edited: May 21, 2015
3. May 20, 2015

### Potatochip911

I believe it would be $arctan(\frac{1}{3})$ although my textbook says "The integral is to be evaluated only at the upper limit; no lower limit is needed" Edit: Well this is embarrassing I was looking at the answer for 5.10 instead of 6.10, whoops

4. May 21, 2015

### SammyS

Staff Emeritus
If you write an integral as a definite integral, it needs both limits.

If you evaluate the indefinite integral at the limit corresponding to infinity, that should be adequate, as your textbook says.

I belive the lower limit for your last integral is $\displaystyle \ arctan\left(\frac{e}{3}\right)$

5. May 21, 2015

### Potatochip911

So is the notation they're using in my textbook wrong then?

6. May 21, 2015

### SammyS

Staff Emeritus
In my opikion, that's not a standard notation.

If your textbook defines it, then it's fine to use in the context of your textbook & class, but I'm not familiar with the notation.