Question

Is there any way to prove that the sum of 1/n! as n -> infinity is convergent other than to compare it with a geometric series?

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 1 }{ n! }\]

Answer 2

The solution I have looks at the denominator first and deduces that\[n! > 2^{n-1}\] and so \[\frac{ 1 }{ n! } < \frac{ 1 }{ 2^{n-1} }\]

Answer 3

\[n!>2^{n-1}\] is not true for any n

Answer 4

there must be a condition where that holds

Answer 5

you would consider the limit of Nth term if it is zero

Answer 6

but that would turn to be difficult, i take it you are looking for a simpler way yes?

Answer 7

this is a set up for the ratio test

Answer 8

check that
\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}<1\]

Answer 9

in fact this limit is zero, since the ratio is
\[\frac{n!}{(n+1)!}=\frac{1}{n+1}\]

Answer 10

thank yooou =)

Answer 11

yw

Answer 12

Another way: Suppose we know that \(e^x\) can be represented as the following series:
\[e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\]
Setting \(x=1\) gives you a finite sum of \(e\), so the series must converge. This assumes you can establish the series converges in the first place, though.