Direct comparison test for convergence

[RadiationX]

I'm supposed to compare the series

\sum_{n=0}^{\infty}\frac{1}{n!}

to some other series to see if the one above converges or diverges. I have no idea of what to compare it to.

I know by the ratio test that the above series converges, that is if I'm doing the ratio test correctly.

\lim_{n\rightarrow\infty}\frac{1}{(n+1)!}n!\\=\frac{1}{n(n+1)}(1*n)=\\\frac{1}{(n+1)}

since this limit is zero which is less than one the series converges

 

[RadiationX]

I'm thinking about majoring in mathematics. How important are sequences and series to a math major? I hate them BTW.

 

[StatusX]

Compare it to the sum of 1/2^n. If you can show this converges, and show that every term in this series is greater than the corresponding term in the original series, that one must also converge.

 

[HallsofIvy]

Or even, and simpler, just compare it to \frac{1}{n^2}. As soon as n> 2,
\frac{1}{n^n}< \frac{1}{n^2}. Now, what do you know about the convergence of \Sigma_1^\infty\frac{1}{n^2}?