<s>[INST] 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 
$$\lim_{n \to \infty} \frac {n} {n}=1$$
 [/INST] 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.)
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