Comparison and Limit Comparison Tests for Convergence of Series

[greenteacup]

Homework Statement

\sum^{\infty}_{n=1} \frac{e^{n}+n}{e^{2n}-n^{2}}

Homework Equations

I have to use either the Comparison Test or the Limit Comparison Test to show whether the series converges or diverges.

The Attempt at a Solution

a_{n} = \frac{e^{n}+n}{e^{2n}-n^{2}}

b_{n} = \frac{1}{e^{2n}}

lim_{n->\infty} \frac{e^{n}+n}{e^{2n}-n^{2}} * e^{2n}

Annnd I'm not sure what to do beyond this point. I'm not even sure I'm taking the right equation for b_{n}... Is it okay to just ignore the e^{n} in the numerator like that?

 

[Office_Shredder]

You can do the limit formally like so

\lim_{n \rightarrow \infty} \frac{e^{3n} + e^{n}n}{e^{2n}-{n^2}}

Divide the top and bottom by e²ⁿ

\lim_{n \rightarrow \infty} \frac{e^n+\frac{n}{e^n}}{1-\frac{n^2}{e^{2n}}}

The numerator goes to infinity as n gets large, but the denominator goes to 1. So obviously the limit doesn't exist. Since the series for b_n converges, you need to find something better.

The n and n² in the limit don't grow nearly as fast as the exponentials, so when considering limit behavior, you might want to assume they're 0 to get a rough idea of what is going on. What can you divide \frac{e^n}{e^{2n}} by to get that the limit as n goes to infinity still exists that might be a suitable candidate?

 

[greenteacup]

Hmm. Could I make it \frac{e^{n}}{ne^{2n}}? Or \frac{ln(e^{n})}{ln(e^{2n})}?

 

[zcd]

Try the limit comparison test with \sum_{n=1}^{\infty}\frac{1}{e^{n}-n}
(this converges if you compare it with the geometric series by the way)

 

[greenteacup]

Thank you so much, everyone! zcd, is the b_{n} you gave me less than \frac{1}{e^{n}} though? It seems like it should be bigger because the denominator is less...

 

[Dick]

Thank you so much, everyone! zcd, is the b_{n} you gave me less than \frac{1}{e^{n}} though? It seems like it should be bigger because the denominator is less...

It is bigger. But it's less than 1/((e^n)/2), for example, since (e^n-n)>((e^n)/2) if n is large enough.