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

In summary, the conversation discusses using the integral test to determine if the series $$ \sum_{n=1}^{\infty} \frac{e^n}{e^{2n}+9} $$ diverges or converges. The attempt at a solution involves rewriting the series and using the integral test to evaluate the integral. The lower limit for the integral is debated, but it is ultimately determined that the series converges. There is also a discussion about notation used in the textbook.
  • #1
Potatochip911
318
3

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?
 
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  • #2
Potatochip911 said:

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.
 
Last edited:
  • #3
SammyS said:
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
 
  • #4
Potatochip911 said:
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) ##
 
  • #5
SammyS said:
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?
Capture.jpg
 
  • #6
Potatochip911 said:
So what the notation they're using in my textbook is wrong then?
Capture.jpg
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.
 

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

1. What is the purpose of the Integral Test for e^n/(e^(2n)+9)?

The Integral Test is used to determine the convergence or divergence of a series by comparing it to an appropriate integral.

2. How is the Integral Test applied to the series e^n/(e^(2n)+9)?

The Integral Test states that if the integral of the series is convergent, then the series is also convergent. In this case, the integral to be evaluated is ∫ e^n/(e^(2n)+9) dn.

3. What is the general formula for the integral of e^n/(e^(2n)+9)?

The integral of e^n/(e^(2n)+9) is given by ∫ e^n/(e^(2n)+9) dn = 1/2ln(e^(2n)+9) + C.

4. How can the convergence of the series e^n/(e^(2n)+9) be determined using the Integral Test?

If the integral ∫ e^n/(e^(2n)+9) dn is convergent, then the series e^n/(e^(2n)+9) is also convergent. If the integral is divergent, then the series is also divergent.

5. Can the Integral Test be used to determine the exact sum of the series e^n/(e^(2n)+9)?

No, the Integral Test only determines the convergence or divergence of a series, it does not provide an exact value for the sum of the series.

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