# lim_(n->oo)(sum_(k=1)^n(sqrt(2n^2+k)/(2n^2+k)))?

Apr 1, 2018

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = \frac{1}{\sqrt{2}}$

#### Explanation:

We have

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{\sqrt{2 {n}^{2} + k}} =$

= lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ represents the $\textcolor{red}{\text{net area}}$ defined by the graph of the function $f \left(x\right)$ and the x-axis. One way to find the value of an integral is by aproximating it with infinitely many rectangles ("boxes") of equal or not width.

This concept is called a $\textcolor{red}{\text{Riemann Sum}}$.

There are infinitely many rectangles, therefore, if there are $n$ rectangles, we have to take the limit as $n \to \infty$.

Now, let's take the case where the width is constant between all boxes. If we define ${x}_{1} , {x}_{2} , \ldots , {x}_{n}$ to be some "marks" on the x-axis such that the width of the $i$-th rectangle is ${x}_{i + 1} - {x}_{i} = \Delta {x}_{i}$, then its lenght is $f \left({x}_{i}\right)$. Since the width is equal, $\Delta {x}_{i}$ is the same for all $i$'s. We can more easily call it color(red)(Delta.

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where color(red)(a = 0 and color(red)(b=1. We have:

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \Delta \cdot f \left({x}_{i}\right)$

This is only true if ${x}_{n}$ is equal to $1$.

One way to make $\Delta$ constant is to define ${x}_{k} = \frac{k}{n}$, for some $k$, $1 \le k \le n$. This way, ${x}_{n} = 1$ and

$\textcolor{red}{\Delta} = {x}_{i + 1} - {x}_{i} = \frac{i + 1}{n} - \frac{i}{n} = \textcolor{red}{\frac{1}{n}}$.

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$.

This looks familiar to our original sum, ${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}}$.

All we are left to do is to assume they're equal and then solve the integral.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$

$i$ and $k$ are just the indexes, the name of which doesn't matter. Both sums could be counting $i$ or $k$, it's irrelevant.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot f \left(\frac{k}{n}\right)$

$\implies f \left(\frac{k}{n}\right) = \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} \implies f \left(x\right) = \frac{1}{\sqrt{2 + \frac{x}{n}}}$.

We have to find the integral of this function:

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2 + \frac{x}{n}}}$

Since $n \to \infty$, this means that $\frac{x}{n} = 0$, as $x$ is finite.

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.

This is because it forms a rectangle with the x-axis, the width and lenght of which is $1$ and $\frac{1}{\sqrt{2}}$, respectively.

As a conclusion,

color(blue)(lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2.