Why does the integral test work when arctan(∞) = nπ/2 for odd n?

[Fernando Rios]

Homework Statement

Use the integral test to find whether the following series converge or diverge.

Relevant Equations

∑ n=1 ∞ (e^n/(e^(2n)+9))

After evaluating the integral I found the following:

(1/3)tan^-1(e^∞/3) = (1/3)tan^-1(∞) = (1/3)(nπ/2), where n is an odd number. In this case I found multiple solutions to the problem. How do you prove it converges?

 

[LCKurtz]

Hopefully, you aren't confusing the n in your answer with the n in the series. Just use $n=1$ in your answer and you are done.

 

[vela]

Remember there's a lower limit on the integral as well, so you don't actually end up with multiple values for the definite integral.

 

[Eclair_de_XII]

Homework Statement: Use the integral test to find whether the following series converge or diverge.
Homework Equations: ∑ n=1 ∞ (e^n/(e^(2n)+9))

After evaluating the integral I found the following:

(1/3)tan-1(e∞/3) = (1/3)tan-1(∞) = (1/3)(nπ/2)

This is very much irrelevant to your conclusion, but isn't there supposed to be another term in your answer? Your infinite series starts at $n=1$. So if each term in your summation is of the form $f(n)=(e^n+9e^{-n})^{-1}$, you'd have to be integrating the function $f(x)=(e^x+9e^{-x})^{-1}$ over the interval $(1,\infty)$, which should produce two terms. And as I recall, the standard $\arctan$ function is bounded between $(-\pi/2,\pi/2)$, so I cannot see how $n$ cannot be anything other than $n=1$.