L-12

\(\displaystyle 12).\;\;\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{2 n^{2}-1}}+\frac{1}{\sqrt{2 n^{2}-2}} +\cdots +\frac{1}{\sqrt{2 n^{2}-n}}\right] \)

SLOUTION :

Above Expression can be written as

\(\displaystyle  \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}-k}}\)

Using Sandwich theorem, we will get 

\(\displaystyle \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}+n}} < \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}-k}} < \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}}}\)

\(\displaystyle \frac{n}{\sqrt{2 n^{2}+n}} < \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}-k}}<\frac{n}{\sqrt{2} \times n}\)

\(\displaystyle \lim _{n \rightarrow \infty} \frac{n}{n \sqrt{2+\frac{1}{n}}} < \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}-k}}<\lim _{n \rightarrow \infty} \frac{1}{\sqrt{2}}\)

\(\displaystyle \frac{1}{\sqrt{2}}<\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{2 n^{2}-k}}<\frac{1}{\sqrt{2}}\)

ANSWER : \(\displaystyle \frac{1}{\sqrt{2}}\)