Find $\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{2n}\right)\ldots\left(1+\dfrac{1}{2^{n-1}n}\right)$ Find $\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{2n}\right)\left(1+\dfrac{1}{4n}\right)\ldots\left(1+\dfrac{1}{2^{n-1}n}\right)$.
(This is not my homework. One of my friends gave this to me.)
 A: Apply the GM-AM inquality to get:
$$\begin{align}
\prod_{j=1}^n\left(1+\frac1{2^{j-1}n}\right)&\leq\left(1+\frac1{n^2}\frac{1+\cdots+2^{n-1}}{2^{n-1}}\right)^n\\
&\leq\left(1+\frac2{n^2}\right)^n
\end{align}$$
Since last expression tends to $1$, the limit is $\leq1$
Since each factor is greater than $1$, the limit must be $1$.
A: Let $y$ be the value of limit. Then,
$$\ln y=\lim_{n\rightarrow \infty} \sum_{k=1}^{n} \ln\left(1+\frac{1}{2^{k-1}n}\right)$$
Since the denominator in $\dfrac{1}{2^{k-1}n}$ is very large, we can work under the approximation $$\ln\left(1+\frac{1}{2^{k-1}n}\right) \approx \frac{1}{2^{k-1}n}$$
Hence,
$$\ln y=\lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{2^{k-1}n}=\lim_{n\rightarrow \infty} \frac{2}{n}\left(1-\frac{1}{2^n}\right)=0$$
$$\Rightarrow y=e^0=\boxed{1}$$
A: Isn't it $1$? The fractional expressions inside the parentheses are zero as $n\to \infty$. 
$$\lim\limits_{n\to\infty}\frac{1}{n}=0$$
$$\lim\limits_{n\to\infty}\frac{1}{2n}=0$$
$$\vdots$$
$$\lim\limits_{n\to\infty}\frac{1}{2^{n-1}n}=0$$
Consequently the answer is $1$ 
