More fun in 2015, Part 26

Calculus Level 4

$L=2016\lim_{n\to\infty}\left(n^{-2016/2015}\sum_{k=1}^{n}\sqrt[2015]{2k-1}\right)$ Find the integer nearest to $L$ .

The answer is 2016.

2 solutions

Otto Bretscher
Oct 29, 2015

$n^{-2016/2015}\sum_{k=1}^{n}\sqrt[2015]{2k-1}$ is the midpoint Riemann sum $\sum_{k=1}^{n}f(x_k)\Delta {x}$ of $f(x)=\sqrt[2015]{2x}$ on the interval $[0,1]$ with $n$ subintervals, with mesh size $\Delta{x}=\frac{1}{n}$ and $x_k=\frac{k-1/2}{n}$ .

Thus $L=2016\sqrt[2015]{2}\int_{0}^{1}\sqrt[2015]{x}dx=2015\sqrt[2015]{2}\approx \boxed{2016}$

Moderator note:

Riemann sums is a good way of dealing with non-polynomial/trigonometric summations that we don't have a simple telescoping approach of working through.

Oh of course! MIDDLE Riemann Sum! The $-1$ term was blocking my work and I thought Riemann Sum is not approachable.

Instead of $2k-1$ , is there a way to solve for $3k-1$ with Riemann Sum as well?

Pi Han Goh - 5 years, 7 months ago

sure... you can make a Riemann sum with $x_k=\frac{k-1/3}{n}$ and consider $f(x)=\sqrt[2015]{3x}$

Otto Bretscher - 5 years, 7 months ago

But this doesn't look like a MIDDLE Riemann Sum. Right? What's the limits in the integral?

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – In a Riemann sum, we can pick the point $x_k$ anywhere on the $k$ th interval ... here we choose it 2/3 along the way.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – Really? You can do that? So the integral is still $\int_0^1 x^{1/2015} \, dx$ ?

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – Yes, indeed. Think about it: All these Riemann sums are squeezed between the upper and the lower sum, so, they all have the same limit.

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – Oh! That's a nice interpretation of Riemann Sum! I didn't consider that thought! Thank you!!!

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – While a agree with what you guys are saying, IMO thinking of it in terms of the "2/3 point Riemann sum" complicates the matter. It is better instead to translate the function and use $g(x) = f(3x-2)$

Calvin Lin Staff - 5 years, 7 months ago

@Pi Han Goh – I want to thank you for the interesting dialogue and for your valiant efforts to teach me (about Otto Stolz etc.).

Otto Bretscher - 5 years, 7 months ago

@Otto Bretscher – NO PROBLEM sir! You've taught me way much more on roots of unity.

Pi Han Goh - 5 years, 7 months ago

Pi Han Goh
Oct 29, 2015

Stolz Lemma can handle this.

Let $\displaystyle a_n = \sum_{k=1}^\infty \sqrt[2015]{2k-1}$ and $b_n = n^{2016/2015}$ .

So $a_n - a_{n-1} = (2n-1)^{1/2015}$ and $b_n - b_{n-1} = n^{2016/2015} - (n-1)^{2016/2015} = \dfrac{2016}{2015}n^{1/2015} + \ldots$ .

We evaluate the limit:

$\begin{aligned} \lim_{n\to\infty} \dfrac {a_n-a_{n-1}}{b_n - b_{n-1}} &=& \lim_{n\to\infty} \dfrac{(2n-1)^{1/2015}}{\frac{2016}{2015}n^{1/2015} + \ldots} \\ &=& \displaystyle 2^{1/2015} \lim_{n\to\infty} \dfrac{n^{1/2015}}{\frac{2016}{2015}n^{1/2015} + \ldots} \\ &=& 2^{1/2015} \cdot \dfrac{2015}{2016} \end{aligned}$

which is a finite number, thus Stolz lemma is applicable. Thus the limit in questions is equal to $2016\cdot 2^{1/2015} \cdot \dfrac{2015}{2016} = 2^{1/2015}\cdot 2015$ .

Now, I'm going to determine whether $2^{1/2015}\cdot 2015$ is closer to 2015 or closer to 2016.

We find the cut-off number: $2015.5$ . Upon division, $2015.5 \div 2015 \approx 1.000248$ . So it's sufficient to prove that if $2^{1/2015} > 1.0002$ , then the desired answer is 2016.

By Bernoulli Expansion, $(1+x)^n \approx 1 + nx$ for small $n$ . In this case, $x = 1, n=\frac1{2015}$ . Upon substitution, we can see that $2^{1/2015}$ is indeed larger than $1.0002$ , thus the answer is $\boxed{2016}$ .