JEE Advanced 2013 · Paper 2, Mathematics question 45

Asymptotics (for $a>-1$): numerator $\sim\dfrac{n^{a+1}}{a+1}$; the bracket is $n^2a+\frac{n(n+1)}2\sim n^2\left(a+\frac12\right)$, so the denominator $\sim n^{a+1}\left(a+\frac12\right)$.

Limit. $\dfrac1{(a+1)\left(a+\frac12\right)}=\dfrac1{60}$ gives $2a^2+3a-119=0$: $a=7$ or $-\frac{17}2$.

Reject $a=-\frac{17}2$: then $\sum k^a$ converges while the denominator grows, so the limit is 0, not $\frac1{60}$.

(B) 7