# evaluating $\lim_{n \to \infty }(\frac{n!}{n^n})^\frac{2n^4 + 1}{5n^5 + 1}$

I was trying to work out the following limit $$\lim_{n \to \infty }\Big(\frac{n!}{n^n}\Big)^{\Large\frac{2n^4+1}{5n^5+1}}$$ I have some idea about how to do it. Basically I expanded the $n!$ and divided each term with n to get $$\lim_{n \to \infty }\Big(\prod_{r=0}^{n-1}\Big(\frac{n-r}{n}\Big)\Big)^{\Large\frac{2n^4+1}{5n^5+1}}$$ and If I take logarithm then I can converte the summation into an integral but I have no clue how to go about treating the $\dfrac{2n^4+1}{5n^5+1}$ part which appears as coefficient in the summation.

• Take $n^5$ common from both num and denom. – Mann Jun 7 '15 at 7:52

You may take the logarithm of your product to get $$\lim_{n \to +\infty}\frac{2n^5 + n}{5n^5 + 1}\times \lim_{n \to +\infty}\frac1n \sum_{k=0}^{n-1}\ln\left(1-\frac{k}{n}\right)$$ giving, with the use of the Riemann sum, $$\frac25 \times \int_0^1 \ln (1-x) dx=-\frac25.$$ as the limit, thus, exponentiating, your desired limit is

$$\color{red}{e^{-2/5}}.$$

• That's a convergent improper integral. How do you know those Riemann sums converge to its value? – zhw. Jun 7 '15 at 21:30
• PS: You have $\ln 0= -\infty$ in your sum. – zhw. Jun 7 '15 at 21:37
• @zhw. Here, assume $x \in [0,1)$, then $x \mapsto \ln(1-x)$ is decreasing giving, for $0\leq k\leq n-2$, $\displaystyle \frac{1}{n}\ln\left(1-\frac{k+1}{n}\right)\leq \int_{k/n}^{(k+1)/n} \ln (1-x)dx \leq \frac{1}{n}\ln\left(1-\frac{k}{n}\right)$ and summing we get that $\displaystyle \frac1n \sum_{k=0}^{n-1}\ln\left(1-\frac{1}{n}\right)$ converges to $\int_0^1 \ln (1-x)dx$, as $n \to +\infty$. And thanks for typo :) – Olivier Oloa Jun 7 '15 at 23:58
• Right. I sometimes wonder if students realize there is something to worry about in such cases, so I bring it up from time to time. PS: You left out $(1/n)$ on the left and right in your inequality. – zhw. Jun 8 '15 at 0:06
• I dislike "mathematicians" who think that "$\lim$" is an operator... and then write lim of thing = lim of other thing = lim of etc... then lim = ... – ParaH2 Jul 13 '15 at 2:15

Use stirlings approximation:

https://en.wikipedia.org/wiki/Stirling%27s_approximation

The limit becomes:

$$(\sqrt{2\pi n}e^{-n})^{\frac{2 n^4+1}{5 n^5+1}}$$

For large values of n, the exponential will be similar to $\frac{2}{5n}$

$$(\sqrt{2\pi n}e^{-n})^{\frac{2}{5 n}}$$

You get

$$(2\pi n)^{\frac{2}{5 n}}e^{-2/5}$$

For large values n, the exponential of the first term will be very close to. It will approach 1 so you end up with:

$$e^{-2/5}$$

We can write the given sequence $a_{n}$ as $$a_{n} = \left\{\left(\frac{n!}{n^{n}}\right)^{1/n}\right\}^{(2 + 1/n^{4})/(5 + 1/n^{5})} = \{b_{n}^{1/n}\}^{c_{n}}\tag{1}$$ where $$b_{n} = \frac{n!}{n^{n}}, \, c_{n} = \frac{2 + (1/n^{4})}{5 + (1/n^{5})}$$ Now we can see that $c_{n} \to 2/5$ as $n \to \infty$ and $$\frac{b_{n + 1}}{b_{n}} = \frac{(n + 1)!}{(n + 1)^{n + 1}}\cdot\frac{n^{n}}{n!} = \dfrac{1}{\left(1 + \dfrac{1}{n}\right)^{n}} \to \frac{1}{e}\text{ as }n \to \infty$$ Hence $b_{n}^{1/n} \to 1/e$. From $(1)$ it is now clear that $a_{n} \to (1/e)^{2/5} = e^{-2/5}$ as $n \to \infty$.

Here we have used the following standard theorem

If $a_{n}$ is a sequence of positive numbers and $a_{n + 1}/a_{n} \to L$ as $n \to \infty$ then $a_{n}^{1/n} \to L$ as $n \to \infty$.

• Nice and Very clever +1. It do not involve any integral nor any Riemann sum – Idris Jun 8 '15 at 4:58