JEE Main 2017 · 9 Apr · Shift 1 · Mathematics Q76

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MathematicsSingle correct+4 / -1Question 76 of 90

If $\displaystyle\lim_{n\to\infty}\frac{1^a+2^a+\dots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\dots+(na+n)]}=\frac1{60}$ for some positive real number a, then a is equal to :

\dfrac{15}2

\dfrac{17}2

Answer and worked solution

Official answer: A

The numerator is about $\dfrac{n^{a+1}}{a+1}$ . The bracket is $n^2a+\frac{n(n+1)}2\approx n^2\frac{2a+1}2$ , so the denominator is about $n^{a+1}\frac{2a+1}2$ . The limit is
$\frac2{(a+1)(2a+1)}=\frac1{60},$
so $2a^2+3a-119=0$ and $a=7$ .

Topics:Limits Special series sums

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JEE Main 2017 · 9 Apr · Shift 1, Mathematics question 76