Question

The value of the limit
$lim _{n→∞}(\frac{1^4+2^4+…+n^4}{n^5}+\frac{1}{√n}(\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+...+\frac{1}{\sqrt{4n}}))$
is equal to___(Rounded off two decimal places)

Answer 1

The correct answer is: 2.19 to 2.21 (approx)