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.