How do I calculate $\lim_{n \to \infty} n^\frac{1}{n} (n+1)^\frac{1}{n+1} ...... (2n)^\frac{1}{2n}$ How do I calculate the limit of the following sequence?
 $$\lim_{n \to \infty} n^{\frac{1}{n}} (n+1)^\frac{1}{n+1} ...... (2n)^\frac{1}{2n}$$
 A: The logarithm of the limiting quantity is: 
$\displaystyle\sum_{k = 0}^{n}\dfrac{1}{n+k}\ln(n+k)$ $= \displaystyle\dfrac{1}{n}\sum_{k = 0}^{n}\dfrac{1}{1+\tfrac{k}{n}}\left(\ln n + \ln\left(1+\tfrac{k}{n}\right)\right) $ $= \displaystyle\dfrac{\ln n}{n}\sum_{k = 0}^{n}\dfrac{1}{1+\tfrac{k}{n}} + \dfrac{1}{n}\sum_{k = 0}^{n}\dfrac{\ln\left(1+\tfrac{k}{n}\right)}{1+\tfrac{k}{n}}$
The second term is a Riemann sum for $\displaystyle\int_{0}^{1}\dfrac{\ln(1+x)}{1+x}\,dx$. 
The first term is $\ln n$ multiplied by a Riemann sum for $\displaystyle\int_{0}^{1}\dfrac{1}{1+x}\,dx$. 
That should be enough to tell you what the limit is.
A: Each factor $(n + k)^{1 \over n + k}$ is at least $n^{1 \over 2n}$ since $n + k \geq n$ and ${1 \over n + k} \geq {1 \over 2n}$. More rigorously you might write
$$(n + k)^{1 \over n + k} \geq n^{1 \over n + k} \geq n^{1 \over 2n}$$
So the product of all $n + 1$ factors is at least $(n^{1 \over 2n})^{n+1} > (n^{1 \over 2n})^n = n^{ 1 \over 2}$. This goes to infinity as $n$ goes to infinity, and therefore your product does too.
A: First, show that:
$$m^{1/m}=e^{\log(m)/m}\geq 1+\frac{\log m}{m}$$
So:
$$\begin{align}
\prod_{m=n}^{2n} m^{1/m}&\geq \prod_{m=n}^{2n}\left(1+\frac{\log m}{m}\right)\\
&\geq 1+\sum_{m=n}^{2n}\frac{\log m}{m}\\
&\geq 1+\int_n^{2n}\frac{\log x}{x}dx\\
&=1+\frac12\left(\log^2(2n)-\log^2(n)\right)\\
&=1+\frac{1}{2}\left(\log(2n)-\log(n)\right)\left(\log(2n)+\log(n)\right)\\
&=1+\frac{1}{2}\log(2)\left(2\log(n)+\log(2)\right)
\end{align}$$
So, the sequence converges to $+\infty$.
A: Easier way:
First, take the logarithm:
$$\frac{\ln(n)}{n} + \frac{\ln(n+1)}{n+1} + \cdots + \frac{\ln(2n)}{2n}$$
Notice that $f(x) = \frac{\ln(x)}{x}$ is a decreasing function after $x = e$ (i.e, it is eventually decreasing). Thus, for sufficiently large $n$:
$$\frac{\ln(n)}{n} + \frac{\ln(n+1)}{n+1} + \cdots + \frac{\ln(2n)}{2n} > \frac{\ln(2n)}{2n} + \frac{\ln(2n)}{2n} + \cdots + \frac{\ln(2n)}{2n} = (n+1) \cdot \frac{\ln(2n)}{2n}$$
(Replacing each term with $\frac{\ln(2n)}{2n}$ is valid because this is the smallest term in the series, because $f(x)$ is decreasing.)
The limit of $(n+1) \cdot \frac{\ln(2n)}{2n}$ as $n$ approaches $\infty$ is infinity.
Since the logarithm of your original product is infinite, your original product is infinite.
