This was the solutiom posted by OP in this query how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? : for the limit : $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$.
My query is related to integral which he/she had performed , isnt that integral supposed to be done when the whole function limit which we are taking has a form of limit sum as a definite integral ? Why we can ignore the "n" term staying outside ? Its also tending to infinity so will not it effect the result of the integral which is being peformed inside which also depends on "n" ?
