Prove that $\lim_{a \to \infty} \sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = 1$.

The above sum (without the $$\lim$$ notation) is convergent $$\forall a \in \Bbb{N}^+$$, because: $$\sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = \sum_{n=1}^{\infty} \frac{n! n! \dots n!}{n^n n^n \dots n^n} \stackrel{\quad \text{because} \\ \forall n \in \Bbb{N}^+ \ n! \le n^n}{\le} \sum_{n=1}^{\infty} \frac{n!}{n^n} \approx \\ \stackrel{\quad \text{Stirling-} \\ \text{approximation}}{\approx} \sum_{n=1}^{\infty} \frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n} = \sum_{n=1}^{\infty} \frac{\sqrt{2\pi n}}{e^n} = \sqrt{2\pi} \sum_{n=1}^{\infty}\frac{\sqrt{n}}{e^n} = \\ = \sqrt{2\pi} \ Li_{-\frac{1}{2}}\left(\frac{1}{e}\right) \approx 1.7728$$ It is also decreasing $$\forall a \in \Bbb{N}^+$$: $$\sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = \sum_{n=1}^{\infty} \left(\frac{(n!)^{a-1}}{n^{(a-1)n}}\right)\frac{n!}{n^n} \le \sum_{n=1}^{\infty} \frac{(n!)^{a-1}}{n^{(a-1)n}}$$ Because the left side is multiplied by a term that is always $$\le 1$$, namely $$\forall n \in \Bbb{N}^+ \quad \frac{n!}{n^n} \le 1$$ So the limit exists: $$\exists \lim_{a \to \infty} \sum_{n=1}^{\infty} \frac{(n!)^a}{n^{an}} = ? \in \Bbb{R}^+_0$$ I examined the sum in Python and wrote a code to calculate it up to $$a=20$$, each time adding up the sum up to $$n=1000$$:

import math

def sum_term(n,a):
return (pow(math.factorial(n),a))/pow(n,a*n)

for a in range(1,21):
value = 0
for n in range(1,1001):
value += sum_term(n,a)
print("a = " + str(a) + " | lim = " + str(value))


And its output was:

a = 1 | lim = 1.879853862175259
a = 2 | lim = 1.3099287490030924
a = 3 | lim = 1.1368584537249211
a = 4 | lim = 1.065018132743388
a = 5 | lim = 1.0317992491522754
a = 6 | lim = 1.0157461094449747
a = 7 | lim = 1.0078393253936435
a = 8 | lim = 1.0039122029986458
a = 9 | lim = 1.0019544471210995
a = 10 | lim = 1.0009768562327848
a = 11 | lim = 1.000488346517213
a = 12 | lim = 1.0002441551281933
a = 13 | lim = 1.0001220735353726
a = 14 | lim = 1.0000610358724382
a = 15 | lim = 1.0000305177372775
a = 16 | lim = 1.0000152588244295
a = 17 | lim = 1.0000076294023905
a = 18 | lim = 1.0000038146990122
a = 19 | lim = 1.000001907349021
a = 20 | lim = 1.0000009536744026


It is pretty convincing that the limit tends to $$1$$, however, I want to prove this mathematically. Is there a way to do this with perhaps the Squeeze theorem or other methods?

• maybe converging to some $1+\epsilon$ and not to $1$. But a very interesting problem, +1 – gt6989b Apr 5 at 15:45
• If it's possible to switch the limit and the sum (maybe monotone convergence theorem?), probably every term except the $n=1$ one will converge to $0$? – TastyRomeo Apr 5 at 15:46

Hint: Note that $$n!/n^n\le 1/n$$ for all $$n.$$ Thus your sum equals $$1+ R(a),$$ where

$$0

Thus all you need to show is that $$R(a)\to 0$$ as $$a\to \infty.$$

Perhaps here is the approach. First term of your sum is 1, and all others are in $$(0,1)$$. So if we exchange the two limits, $$x_n^a \to 0$$ as $$a \to \infty$$, so the entire sum will be zero.

You have to check the axioms if the limits can be exchanged though, but everything looks safe -- both positive and convergent...

• Switching up the two limits seems to give the result of $0$, but I'm not sure why you wouldn't be able to switch the limits.... hmmm.... – Daniel P Apr 5 at 16:05
• @DanielP the sum is still 1, since none of this applies to the first term... – gt6989b Apr 5 at 17:16

It is elementary that $$\frac{n!}{n^n}\le\frac12$$ for all $$n>1$$ and the series is squeezed in

$$\left[1,\frac1{1-\dfrac1{2^a}}\right].$$

For large $$a$$, is is approximately

$$\dfrac1{1-\dfrac1{2^a}}$$ as witnessed by

$$\dfrac1{1-\dfrac1{2^{20}}}=\color{green}{1.00000095367}522590\cdots$$