Exponent Year

Calculus Level 3

$\lim_{x\rightarrow \infty} \frac{1^{2017}+2^{2017} + 3^{2017}+\ldots+x^{2017}}{\left(x^{1009}+\frac{1}{x^{1009}}\right)^2} = \frac {1}{A}$ . Find $A$ .

The answer is 2018.

1 solution

Arghyadeep Chatterjee
Nov 27, 2019

As $x\to\infty$ we can approximate $(x^{1009} + \frac{1}{x^{1009}})^{2}$ as $(x^{1009})^{2}$

So the problem transforms to:-

$\large \lim_{x\to\infty}\frac{\sum_{r=1}^{\infty}r^{2017}}{x^{2018}}$

$\large \lim_{x\to\infty}\frac{1}{x}\sum_{r=1}^{\infty}(\frac{r}{x})^{2017}$

Using Riemann Sums

We have it as

$\large \int_{0}^{1}x^{2017} dx$

$\large = \frac{1}{2018}$

Exponent Year

The answer is 2018.

1 solution

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