# Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$

I am stuck on this problem:

Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by

$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$

So I am trying to find:

$$\lim_{n \to \infty}\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$

In a situation like this should I be noting that the denominator is increasing in value faster than the numerator?

My first thought was to do some manipulation. I may have done something incorrectly. I began with the following.

$$\frac {n^2} {\sqrt{n^6+k}}=\frac {n^2} {\sqrt{n^6(1+k/n^6)}}=\frac {n^2} {\sqrt{n^6}\sqrt{1+k/n^6}}=\frac {1} {n} \cdot \frac {1} {\sqrt {1+k/n^6}}$$

So I now have

$$\lim_{n \to \infty} \frac {1} {n} \sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}$$

Now, I am unsure about the following. It looks to me as if $k/n^6$ goes to zero as $n \to \infty$. That would result in $\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$. So I would be left with

$$\lim_{n \to \infty} \frac {n} {n}=1$$

-

That looks great!

I'll show you another, slightly cleaner (in my opinion) method:

We seek $\displaystyle \lim_n \sum_{k = 1}^n \frac{n^2}{\sqrt{n^6 + k}}$

Each term in the sum shares bounds, namely $\dfrac{n^2}{\sqrt{n^6 + n}} \leq \dfrac{n^2}{\sqrt{n^6 + k}} < \dfrac{n^2}{\sqrt{n^6}}$

So then $\displaystyle 1 =\dfrac{n^3}{\sqrt{n^6 + n}} \leq \lim_n \sum_{k = 1}^n \frac{n^2}{\sqrt{n^6 + k}} \leq \lim \dfrac{n^3}{\sqrt{n^6}}=1$

-
You beat me by 50 seconds! :P – Daniel Montealegre Mar 8 '12 at 5:07

I think this should work, $$\frac{n\cdot n^2}{\sqrt{n^6+k}}\leq \sum_{k=1}^n\frac{n^2}{\sqrt{n^6+k}}\leq \frac{n\cdot n^2}{\sqrt{n^6+1}}$$Taking limits at both sides says the limit is $1$.

-
Nice answer! :) – Isaac Solomon Mar 14 '12 at 5:25

The conclusion is correct, but the step $$\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$$ is clearly wrong, a fact that the other answers unfortunately failed to mention: each of the $n$ denominators is greater than $1$, so each term of the sum is less than $1$, and the sum itself is less than $n$.

The largest term in the sum is $$\frac1{\sqrt{1+1/n^6}}\;,$$ and the smallest is $$\frac1{\sqrt{1+k/n^6}}\;,$$ so you do know that

$$\frac{n}{\sqrt{1+k/n^6}}\le\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}\le\frac{n}{\sqrt{1+1/n^6}}$$

and hence that

$$\frac1{\sqrt{1+k/n^6}}\le\frac1n\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}\le\frac1{\sqrt{1+1/n^6}}\;.$$

With this in hand you’re in business, since the limit as $n\to\infty$ of each of the bounds is $1$.

(Note that all of the answers work in essentially the same way, by trapping the expression between two simpler ones with the same limit.)

-