Definite Integration Question 214
Question: $ \underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{n}+\frac{1}{\sqrt{n^{2}+n}}+\frac{1}{\sqrt{n^{2}+2n}}+…..+\frac{1}{\sqrt{n^{2}+(n-1)n}} ] $ is equal to
[RPET 2000]
Options:
A) $ 2+2\sqrt{2} $
B) $ 2\sqrt{2}-2 $
C) $ 2\sqrt{2} $
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=\underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{n}+\frac{1}{\sqrt{n^{2}+n}}+….+\frac{1}{\sqrt{n^{2}+(n-1)n}} ] $
therefore $ y=\underset{n\to \infty }{\mathop{\lim }}[ \frac{1}{n}+\frac{1}{n\sqrt{1+\frac{1}{n}}}+….+\frac{1}{n\sqrt{1+\frac{(n-1)}{n}}} ] $
therefore $ y=\frac{1}{n}\underset{n\to \infty }{\mathop{\lim }}[ 1+\frac{1}{\sqrt{1+\frac{1}{n}}}+….+\frac{1}{\sqrt{1+\frac{(n-1)}{n}}} ] $
$ y=\underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}\sum\limits _{k=1}^{n}{\frac{1}{\sqrt{1+\frac{(k-1)}{n}}}} $ , Put $ \frac{k-1}{n}=x $ and $ \frac{1}{n}=dx $
therefore $ y=\underset{n\to \infty }{\mathop{\lim }}\int\limits_0^{\frac{n-1}{n}}{\frac{dx}{\sqrt{1+x}}} $
$ =\underset{n\to \infty }{\mathop{\lim }}2[ \sqrt{1+x} ] _{0}^{( \frac{n-1}{n} )} $
therefore $ y=2\underset{n\to \infty }{\mathop{\lim }}[ \sqrt{\frac{2n-1}{n}}-1 ] $
$ =2\underset{n\to \infty }{\mathop{\lim }}\sqrt{\frac{2n-1}{n}}-2 $
therefore $ y=2\underset{n\to \infty }{\mathop{\lim }}\sqrt{2-\frac{1}{n}}-2 $
$ =2\sqrt{2}-2 $ .
BETA
