$\lim_{n\to \infty} \sqrt[n]{(1+\frac{2}{n})\cdot(1+\frac{4}{n})\cdot(1+\frac{6}{n})\cdot...\cdot(1+\frac{2n}{n})}$ =
$$\lim_{n\to \infty} \left((1+\frac{2}{n})\cdot(1+\frac{4}{n})\cdot(1+\frac{6}{n})\cdot...\cdot(1+\frac{2n}{n})\right)^\frac{1}{n}$$
I would say the result is 1 but I know that is wrong, can someone help me please?
HINT: If we take logarithm of this expression we obtain $$ \frac1n(\ln(1+2/n)+\ln(1+4/n)+\ldots\ln(1+(2n)/n)). $$ Can you see a Riemann sum now?
Edit: Take $f(x)=\ln(1+2x)$, $x\in[0,1]$. Divide interval $[0,1]$ onto $n$ equal subintervalss and take right hand sides of them. Then $$ \frac1n\sum_{k=1}^n\ln\left(1+\frac{2k}{n}\right) $$ is a Riemann sum of $$ \int_0^1\ln(1+2x)\,dx. $$ This integral exists and is easy to compute. From the definition of definite (Riemann) integral it is equal to the limit in question.