Problem in Riemann Summation method 
$$\lim_{n\to\infty}\left\{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\left(1+\frac{4n}{n}\right)\right\}^{1/n}=\frac{5^5}{e^4}$$

I want to prove this using the Riemann summation method (the one in which $r/n$ is $x$ and $1/x$ is $\mathrm{d}x$ and then integral is evaluated with limits $0$ to $1$) I tried but the answer isn't matching the RHS.
My Attempt:

*

*I took the logarithm of both sides (I let RHS = L).

*After conversion and series the integral was $\int_0^1\ln(1+x)\,\mathrm{d}x$

*After calculating and taking antilog my answer was $4/e$

*So where did I make a mistake?

Thanks in advance.
 A: Your problem lies in that you’re blindly converting any such sum into an integral wil limits 0 and 1. In fact, the limits are 0 to 1 ONLY when the value of the variable r finishes at n (starting from 1). If the value goes from 1 to kn, then the integral must have limits 0 to k, not 1.
Let $$L=\lim_{n\to\infty}\left\{\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\left(1+\frac{4n}{n}\right)\right\}^{1/n}$$$$\implies \ln L  =\lim_{n\to \infty} \frac 1n \sum_{r=1}^{4n} \ln\left(1+\frac rn\right)=\int_0^4\ln(1+x)dx$$$$=\int_1^5\ln u   du=(u\ln u-u)\Bigg |_1^5$$$$=(5\ln 5-5)-(1\ln1-1)=5\ln 5-4$$ Thus, $$L=\exp(5\ln 5-4)=\frac{5^5}{e^4}.$$
A: While this has already been sufficiently answered, I also wanted to point out that the original problem can be defined as a special type of integral called a product integral (specifically the geometric integral). Similar to the Riemann sum definition of the integral, the geometric integral can be defined as
$$\prod_a^bf(x)^{dx}=\lim_{n\to\infty}\prod_{i=1}^nf(x_i)^{\Delta x}$$
where $\Delta x=\frac{b-a}n$ and $x_i$ is a value on the closed interval $[a+(i-1)\Delta x,a+i\Delta x]$. It can be shown through the same methods as above that
$$\prod_a^bf(x)^{dx}=\exp\left(\int_a^b\ln f(x)dx\right)$$
Applying this to the original problem we have $f(x)=1+x$ and
$$\prod_0^4(1+x)^{dx}=\exp\left(\int_0^4\ln(1+x)dx\right)$$
$$=\exp\left(\int_1^5\ln udu\right)=\exp\left(u\ln u-u\Bigg |_1^5\right)$$
$$=\exp(5\ln5-4)=\frac{5^5}{e^4}$$
