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.)
Apply the GM-AM inquality to get:

$$\begin{align}
\prod_{j=1}^n\left(1+\frac1{2^{j-1}n}\right)&amp;\leq\left(1+\frac1{n^2}\frac{1+\cdots+2^{n-1}}{2^{n-1}}\right)^n\\
&amp;\leq\left(1+\frac2{n^2}\right)^n
\end{align}$$

Since last expression tends to $1$, the limit is $\leq1$

Since each factor is greater than $1$, the limit must be $1$.
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 $1$