We apply the $\log$ function and we use the Riemann sum and we integrate:
$$\log\left(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}\right)=-\log n+\frac{1}{n}\sum_{k=1}^n\log(k+n)=\frac{1}{n}\sum_{k=1}^n\log(k+n)-\log n\\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_0^1 \log(1+x)dx=2\log(2)-1$$
hence
$$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}=\frac{4}{e}$$
I am sorry for overlooking it earlier, but in the evaluation of $\sum\limits_{i=1}^n f(x_i)\Delta_{x_i} = \sum\limits_{i=1}^n (1+(\frac{5i}{n})^3).(\frac{5}{n}) = 5 + \frac{625}{n^4}(\sum\limits_{i=1}^n i^3)$,
So you wont be needing $\sum\limits_{i=1}^n i^4$ as such.
But if you are still interested in knowing how to find the summation without having to apply mathematical induction I know a fast way of arguing combinatorially for $n>6$,
Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,
then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$
the positions can be filled in $\sum\limits_{i=1}^n i^4$ ways.
Counting it according to the cases: $5 = 4 + 1 = 2 + 2 + 1 = 3 + 1 + 1 = 2+1+1+1=1+1+1+1+1$ (where the numbers in partitions indicate the number of equal components in that partition, i.e. for example $5=4+1$ corresponds to the case $x_1=x_2=x_3=x_4<x_5$; and $2+2+1$ corresponds to the case two pairs have equal value but the value of the pair are different ,.. and so on). Count the number of ways in which each of the partitioning can happen: $1,14,36,24$ respectively (in the same order as I have presented the partitions).
You get the number of ways to be: ${n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5}$
Therefore, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.
Best Answer
We have $$\lim_{n\to\infty}\frac1n\sum_{r=1}^n\sqrt{\dfrac rn}$$
Now, $$\lim_{n\to\infty}\frac1n\sum_{r=1}^nf\left(\dfrac rn\right)=\int_0^1 f(x)\ dx$$