How do I prove: limn→+∞n!n is infinte?

Question

How do I prove:  limn→+∞n!n is infinte?

Joseph Lewis · Accepted Answer

By considering Taylor series, ex≥xnn! for all x≥0, and n∈N. In particular, for x=n this yields  n!≥(ne)n  Thus  n!n≥ne→∞

Cleveland Walters · Accepted Answer

We wish to show that for a fixed real number

α

, we have

{(n!)}^{\frac{1}{n}} > α

for sufficiently large n. Clearly it suffices to show that the logarithm of this quantity exceeds

α

(for sufficiently large n).
Using elementary log rules, we have

\log (n!^{\frac{1}{n}}) = \frac{\log (n!)}{n}

= \frac{1}{n} \sum_{i = 1}^{n} \log (i)

We will start the sum at 2since

\log (1)

is o. Now

\frac{1}{n} \sum_{i = 2}^{n} \log (i)

is a right-handed Riemann sum (with step-size

△ x = 1

) which gives an overestimate for the integral

\int_{1}^{n} \log (x) d x

But this integral is just

n \log (n) - n + 1

Thus we have shown that

\log (n!^{\frac{1}{n}}) \geq \log (n) - 1 + \frac{1}{n}

and the right hand side goes to

\infty

as n goes to

\infty

, which is what we want.

alenahelenash · Accepted Answer

Denote Sn=(nn!). Then log⁡Sn=log⁡1+...+log⁡nn=anbn Since bn is increasing and tends to ∞ and we have an+1−anbn+1−bn=log⁡(n+1)→∞ we can apply Stolz-Cesaro theorem and conclude that the limit L=limn→∞Sn exists and L=∞. This implies that Sn→∞

How do I prove: \lim_{n\to+\infty}\root(n)(n!) is infinte?

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