$$\left[\lim _{n \rightarrow \infty}\left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{n}}{n^{n+1}}\right)\right)\right]1$$
The answer of this question is 0. But my answer is e. Here is my approach. We can simplify the product inside the limit using the following properties of limits:
The product of limits is equal to the limit of the product (if the individual limits exist). The limit of a sum is equal to the sum of the limits (if the individual limits exist). Applying these properties, we can rewrite the product inside the limit as:
\begin{gathered} \rm\lim_{n\rightarrow\infty} \left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right) \\ \end{gathered}
We can then take the natural logarithm of both sides and use the properties of logarithms to simplify the expression further:
\begin{gathered}\rm\ln\left(\lim_{n\rightarrow\infty} \left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right) \\ \end{gathered}
\begin{gathered} \rm=\lim_{n\rightarrow\infty} \ln\left(\left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right) \\ \end{gathered}
\begin{gathered} \rm=\lim_{n\rightarrow\infty} \left(\ln\left(1+\frac{2^n}{n^{n+1}}\right)+\ln\left(1+\frac{3^n}{n^{n+1}}\right)+\cdots+\ln\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right) \\ \end{gathered}
We can then use the property that for small values of x, ln(1+x) is approximately equal to x, to approximate each term in the sum:
\begin{gathered} \rm\ln\left(1+\frac{k^n}{n^{n+1}}\right) \approx \frac{k^n}{n^{n+1}} \\ \end{gathered}
Using this approximation, we get:
\begin{gathered} \rm\lim_{n\rightarrow\infty} \left(\frac{2^n}{n^{n+1}}+\frac{3^n}{n^{n+1}}+\cdots+\frac{(n+1)^n}{n^{n+1}}\right) \\ \end{gathered}
We can then use the fact that the largest term in the sum dominates as n goes to infinity, to approximate the sum with the last term:
\begin{gathered} \rm lim_{n\rightarrow\infty} \frac{(n+1)^n}{n^{n+1}} = \lim_{n\rightarrow\infty} \left(\frac{n+1}{n}\right)^n = e \\ \end{gathered}
Therefore, the final answer is:
\begin{gathered} \rm \lim_{n\rightarrow\infty} \left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{n}}{n^{n+1}}\right)\right) = e.\\\end{gathered}
We have $$0<\ln\left (1+{k^n\over n^{n+1}}\right )\le {k^n\over n^{n+1}}\quad (*)$$ Next for a fixed $k\le n$ we have $$2^n+3^n+\ldots +(n+1-k)^n+(n+2-k)^n\ldots +(n+1)^n\\ \le n(n+1-k)^n+k(n+1)^n$$ Hence $${2^n+3^n+\ldots +(n+1)^n\over n^{n+1}}\le {(n+1-k)^n\over n^n}+{k\over n}{(n+1)^n\over n^n} \\ =\left (1-{k-1\over n}\right )^n+{k\over n}\left (1+{1\over n}\right )^n$$ We obtain $$\limsup_n {2^n+3^n+\ldots +(n+1)^n\over n^{n+1}}\le e^{1-k}$$ As $k$ is arbitrary, we get $$\lim_n {2^n+3^n+\ldots +(n+1)^n\over n^{n+1}}=0$$ By $(*)$ we get $$\lim_n \ln\left [\prod_{k=1}^n \left (1+{k^n\over n^{n+1}}\right )\right ]=\lim_n \sum_{k=1}^n\ln\left (1+{k^n\over n^{n+1}}\right ) =0$$ Therefore the limit of the product is equal $1.$