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

[Potatochip911]

Homework Statement

Using the integral test determine whether or not the following sum diverges. $$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9} $$

Homework Equations

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?

 

[SammyS]

Homework Statement

Using the integral test determine whether or not the following sum diverges. $$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9} $$

Homework Equations

3. The Attempt at a Solution [/B]
$$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9}=\sum_{n=1}^{\infty} \frac{e^n}{({e^{n})}^2+9} \\
\int^{\infty} \frac{e^n}{({e^{n})}^2+9} \cdot dn; \hspace{0.5cm} Let \space u=e^n; \hspace{0.5cm} du=e^n\cdot dn \\
\int^{\infty} \frac{du}{u^2+9}; \hspace{0.5cm} Let \space u=3\tan\theta \hspace{0.5cm} du=3\sec^2\theta\,\color{red}{d\theta} \\
\int^{(\frac{\pi}{2})^-} \frac{3\sec^2\theta}{9\tan^2\theta+9}\color{red}{d\theta}=\frac{1}{3} \int^{(\frac{\pi}{2})^-} 1\cdot d\theta=\frac{\pi}{6} $$ 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?

What is the lower limit for the integral?

It is true that this series converges.

Added in Edit:

You have left dθ out of a couple of expressions.

 

[Potatochip911]

What is the lower limit for the integral?

It is true that this series converges.

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

 

[SammyS]

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

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

 

[Potatochip911]

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

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

 

[SammyS]

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

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.