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$$a_n = \frac{1}{n^2} \sum _{k=1}^n \left( \sqrt{\left(n+k\right)^2+n+k}\, \right)$$

$$\lim_{n\to \infty} a_n =\:?$$

I solved it like this and I would want to know it this makes sense and if not why.

$$\frac{1}{n}\sum _{k=1}^n \left[ \sqrt{\frac{\left( n^2+2kn+k^2+n+k \right)}{n^2}} \; \right]= \frac{1}{n}\sum _{k=1}^n \left[ \sqrt{1+2\frac{k}{n}+\left( \frac{k}{n} \right)^2+ \frac{1}{n}+\frac{k}{n}\cdot\frac{1}{n}} \; \right]$$

and because $n$ tends to infinity, $\dfrac{1}{n}$ and $\dfrac{k}{n} \times \dfrac{1}{n}$ disappears. But remains

$$a_n=\frac{1}{n}\sum _{k=1}^n \sqrt{\left( \frac{k}{n}+1 \right)^2}$$

And so I considered a function $f:[0, 1] \to \mathbb{R}$, $f(x)=x+1$ and then I integrated for $x$ from $0$ to $1$ and got $\frac{3}{2}$, which should be the limit of $a_n$.

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    You forgot the big radical sign in the second summation expression for $a_n$. – 2017-01-08
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    "...and because $\;n\to\infty\;$ it remains..."?? First, you didn't write the square root in the middle steps, second: The right side of your first line of calculations has no $\;k^2\;$ at all in it... – 2017-01-08
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    @DonAntonio Fixed it. – 2017-01-08
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    @Liviu So if I understood correctly, the missing terms into the square root, which are $\;\cfrac k{n^2}+\cfrac1n\;$ , you just ommited them? – 2017-01-08
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    @DonAntonio Well... i don't understand the concept very well and i thought if n tends to infinity 1/n tends to 0 so i can ommit them. How could i have formed the function if there was $1/n * k/n$ ? – 2017-01-08
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    But you should know that you can't take to the limit **part of the expression** while leaving the rest of it untouched...unless you first prove this is (mathematically) legal ! – 2017-01-08
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    Ok.. so if i leave those terms there what function could i form ? – 2017-01-08
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    @Liviu I really can't say...are you sure this must be done with Riemann sums? – 2017-01-08
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    @DonAntonio Yes, it's classified in Riemann Sums chapter and i did all the other exercises but not this. – 2017-01-08

2 Answers 2

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$$(n+k)^2+(n+k)=(n+k)(n+k+1)$$

For $k=1,2,\ldots,n$

$$(n+k)^2< (n+k)(n+k+1) <(n+k+1)^2$$

Now $$\frac{1}{n^2} \sum_{k=1}^{n} (n+k)

$$ \frac{1}{n^2} \left[ n^2+\frac{n(n+1)}{2} \right]

$$ \frac{3}{2}+\frac{1}{2n}

Take the limits for the upper and lower bounds and by sandwich theorem, we get $$\lim_{n\to \infty} a_{n} = \frac{3}{2}$$

Further points to be noticed

If your sum is running from $k=0$ to $k=n-1$, then your sum is exactly between the left and right rectangular rules of $\displaystyle \int_{0}^{1} (x+1)\, dx$.

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    So what i did was ok, right ? I got to the same answer, now i realise that when i calculated $\int _0^1\left(x+1\right)dx\:$ i didn't integrate first (dumb mistake) but it was also $\frac{3}{2}$. – 2017-01-08
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    I recommend to write your summand as $ \sqrt{\left( \dfrac{n+k}{n} \right)^2+O\left( \dfrac{1}{n} \right)}=1+\dfrac{k}{n}+O\left( \dfrac{1}{n} \right)$ first. – 2017-01-08
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The difference between $\sqrt{(n+k)^2+(n+k)}$ and $\left(n+k+\tfrac{1}{2}\right)$ is pretty small: $$ 0\leq \left(n+k+\tfrac{1}{2}\right)-\sqrt{(n+k)^2+(n+k)} \leq \frac{1}{4(n+k)} $$ hence the limit stays the same by replacing $\sqrt{(n+k)^2+(n+k)}$ with $\left(n+k+\tfrac{1}{2}\right)$.
In the latter case it is: $$ \lim_{n\to +\infty}\frac{3n^2+2n}{2n^2} = \color{red}{\frac{3}{2}}.$$