Question

$\displaystyle\rm \red {\left[\lim _{n \rightarrow \infty}\left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{n}}{n^{n+1}}\right)\right)\right]}$​

Answer 1

Answer:

Step-by-step explanation:

Answer is given in picture

Answer 2

$\displaystyle\rm \red {\left[\lim _{n \rightarrow \infty}\left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{n}}{n^{n+1}}\right)\right)\right]}$

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:

$\rm\lim_{n\rightarrow\infty} \left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right)$

We can then take the natural logarithm of both sides and use the properties of logarithms to simplify the expression further:

$\rm\ln\left(\lim_{n\rightarrow\infty} \left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right)$

$\rm=\lim_{n\rightarrow\infty} \ln\left(\left(1+\frac{2^n}{n^{n+1}}\right)\left(1+\frac{3^n}{n^{n+1}}\right)\cdots\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right)$

$\rm=\lim_{n\rightarrow\infty} \left(\ln\left(1+\frac{2^n}{n^{n+1}}\right)+\ln\left(1+\frac{3^n}{n^{n+1}}\right)+\cdots+\ln\left(1+\frac{(n+1)^n}{n^{n+1}}\right)\right)$

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:

$\rm\ln\left(1+\frac{k^n}{n^{n+1}}\right) \approx \frac{k^n}{n^{n+1}}$

Using this approximation, we get:

$\rm\lim_{n\rightarrow\infty} \left(\frac{2^n}{n^{n+1}}+\frac{3^n}{n^{n+1}}+\cdots+\frac{(n+1)^n}{n^{n+1}}\right)$

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:

$\rm lim_{n\rightarrow\infty} \frac{(n+1)^n}{n^{n+1}} = \lim_{n\rightarrow\infty} \left(\frac{n+1}{n}\right)^n = e$

Therefore, the final answer is:

$\rm \lim_{n\rightarrow\infty} \left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{n}}{n^{n+1}}\right)\right) = e.$