JEE Advanced 2017 Paper 2 Mathematics: Paragraph 2

From $a_{n+1}=\dfrac{a_n}{1+na_n}$ we get $b_{n+1}=\dfrac{1}{a_{n+1}}=\dfrac{1+na_n}{a_n}=b_n+n$. Telescoping from $b_1=1$: $b_n=1+\sum_{k=1}^{n-1}k=1+\dfrac{n(n-1)}{2}$, so $a_n=\dfrac{2}{n^2-n+2}$. The condition $a_n<\frac1{1000}$ means $1+\frac{n(n-1)}2>1000$, i.e. $n(n-1)>1998$. Since $45\cdot44=1980<1998$ but $46\cdot45=2070>1998$, the smallest such $n$ is $46$. Recognising the telescoping structure of the reciprocals is the entire trick, so b_n is just one plus a triangular number. Checking neighbouring values avoids off-by-one errors: a_45 equals 2/1982, which exceeds 1/1000, whereas a_46 equals 2/2072 and is below it.

We have $a_k=\dfrac{2}{k^2-k+2}$, which behaves like $\dfrac2{k^2}$ for large $k$. Then $\sum_{k=n}^{2n}a_k\approx\int_n^{2n}\dfrac{2}{x^2}\,dx=2\left(\dfrac1n-\dfrac1{2n}\right)=\dfrac1n$. More rigorously, $n\sum_{k=n}^{2n}a_k=\dfrac1n\sum_{k=n}^{2n}\dfrac{2}{(k/n)^2-k/n^2+2/n^2}$, which is a Riemann sum in $u=k/n$ and tends to $\displaystyle\int_1^2\dfrac{2}{u^2}\,du=1$. Hence the limit equals $1$. Numerical evaluation with the exact sequence for $n=1000$ gave $1.002$, trending to $1$, which supports the answer; $2$ would arise from forgetting the upper limit $2n$. The scale factor n in front of the sum is exactly what converts the sum of order 1/n into a finite limit, so the answer is the definite integral of 2/u squared from 1 to 2.