Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
up vote 6 down vote accepted

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

share|cite|improve this answer
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 $1$.

share|cite|improve this answer
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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.