Convergence of 1/K!

1. Apr 15, 2015

RyanTAsher

1. The problem statement, all variables and given/known data

Does (1/(k!)) converge?

2. Relevant equations

Convergence Tests?

3. The attempt at a solution

I thought I could just simply use the divergence test, but I'm not sure if that only tells you if it's divergent and not whether it is convergent or not.

lim(k>inf) (1/(k!)) = 0, if L =/= 0, then it is divergent, but since it is 0 does that make it convergent, or inconclusive?

I've thought about trying a comparison test but I'm not sure what I could compare 1/k! too, would I compare it to 1/k? In which case it would be divergent.

Kinda stuck here on which to use.

2. Apr 15, 2015

StateOfTheEqn

Are you wondering about the convergence of the series $\Sigma_{k=0}^\infty \frac{1}{k!}$

We can set $e^x=\Sigma_{k=0}^\infty \frac{x^k}{k!}$. Then $e^1=\Sigma_{k=0}^\infty \frac{1^k}{k!}=\Sigma_{k=0}^\infty \frac{1}{k!}$

So, the series converges and equals $e$.

3. Apr 15, 2015

HallsofIvy

Staff Emeritus
While it is true that $\sum \frac{1}{k!}$ converges to e, if you did not know that initially, you could have used the "ratio test": $\sum a_n$ converges if $\frac{a_{n+1}}{a_n}< 1$ and diverges if it is greater than one.

(Your test, that if $\sum a_n$ converges, the $a_n$, must go to 0, is generally called the "divergence test" since it tells us that if that limit is NOT 0 then the series diverges but does NOT tell us one way or the other if the limit is 0.)

4. Apr 15, 2015

phyzguy

I don't think you stated the ratio test quite right. The ratio test tells us that the series converges if $lim_{n->\infty} \frac{a_{n+1}}{a_n} <1$. The series the OP asked about certainly passes this test, since $lim_{n->\infty} \frac{n!}{(n+1)!} = lim_{n->\infty} \frac{1}{n+1} = 0$. You wrote that the ratio test is that the series converges if $\frac{a_{n+1}}{a_n} < 1$. The series $\sum_{n=1}^\infty \frac{1}{n}$ would pass this test since $\frac{a_{n+1}}{a_n} = \frac{n}{n+1} < 1$ for all n, but it does not converge. The ratio test for this series gives $lim_{n->\infty} \frac{a_{n+1}}{a_n} = lim_{n->\infty} \frac{n}{n+1} = 1$