$\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)=$

$\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n} = $

$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n}\right]=$

$\lim _{n \rightarrow \infty} \frac{(2n(2n-1) \dots (n+1))^{1/n}}{n} = $

$a \in \mathbb{R}$ (બધી વાસ્તવિક સંખ્યાઓનો ગણ) માટે,$a \neq -1$,જો $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$ હોય,તો $a$ ની કિંમત શોધો:

જો $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$ હોય,તો $\log k=$