Convergence of 1/n!

[fred2028]

How would you mathematically find the convergence of

\(\displaystyle \[\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}} \]\)

?

By logic, I assume it converges since as n increases, the term will decrease since each term will be 1 divided by a bigger #. However, how would I show this to be true, or false, mathematically?

 

[BigGlenntheHeavy]

Use ratio test, converges absolutely to e.

 

[galactus]

As Glenn pointed out, the series converges to e. Know how to show that?.

Just use the Taylor series for e and let x=1.

\(\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\)

Therefore, when x=1, we have \(\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}=e^{1}\)

 

[Deleted member 4993]

How would you mathematically find the convergence of

\(\displaystyle \[\sum\limits_{n = 0}^\infty {\frac{1}{{n!}}} \]\)

?

By logic, I assume it converges since as n increases, the term will decrease since each term will be 1 divided by a bigger #. However, how would I show this to be true, or false, mathematically?

That is a FALSE logic - the harmonic series ?(1/n) diverges.

Lim (n ? ?)(a[sub:4dsy5dx1]n[/sub:4dsy5dx1]) ? 0 is a necessary condition - not sufficient.