Applying squeeze theorem to a given function

Question

How am I supposed to squeeze a given function between two other functions while applying "squeeze theorem"?

For instance, take this example:

Q.Evaluate: $$\lim _{n\to \infty \:}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}\right)$$

Let us say, the function inside the bracket is $g(n)$. Now how can I find two other functions that such that $f(n)≤ g(n)≤ h(n)$.

Answer to this question is $1$.

Answer 1

As , $$n^2+1 \le n^2+1 \le n^2+n , \\ n^2+1 \le n^2+2 \le n^2+n, \\ n^2+1 \le n^2+3 \le n^2+n, \\ \vdots \\ n^2+1 \le n^2+n \le n^2+n $$ $\implies$ $$ \frac{1}{n^2+n} \le \frac{1}{n^2+1} \le \frac{1}{n^2+1} ,\\ \frac{1}{n^2+n} \le \frac{1}{n^2+2} \le \frac{1}{n^2+1} , \\ \frac{1}{n^2+n} \le \frac{1}{n^2+3} \le \frac{1}{n^2+1} ,\\ \vdots \\ \frac{1}{n^2+n} \le \frac{1}{n^2+n} \le \frac{1}{n^2+1}$$

(As, if $\displaystyle a\le b \le c \implies a\le b , b\le c \implies \frac{1}{b} \le \frac{1}{a} , \frac{1}{c} \le \frac{1}{b} \implies \frac{1}{c} \le \frac{1}{b} \le \frac{1}{a} $)

Multipling $n$ to each inequality and summing them,we have,

$$\left(\frac{n}{n^2+n}+\frac{n}{n^2+n}+...+\frac{n}{n^2+n}\right) \le \left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots +\frac{n}{n^2+n}\right) \le \left(\frac{n}{n^2+1}+\frac{n}{n^2+1}+ \cdots +\frac{n}{n^2+1} \right) $$ $\implies $ $$ \frac{n^2}{n^2+n} \le \left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots +\frac{n}{n^2+n}\right) \le \frac{n^2}{n^2+1} $$ $\implies $ $$ \frac{1}{1+\frac{1}{n}} \le \left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots +\frac{n}{n^2+n}\right) \le \frac{1}{1+\frac{1}{n^2}} $$

Answer 2

A useful tool that is often used is that when $a\leq b$, then $\frac{1}{a} \geq \frac{1}{b}$. So in this case $n^2\leq n^2+k$ for all $k \in \mathbb{N}$. So we have $\frac{1}{n^2} \geq \frac{1}{n^2+k}$ for all $k \in \mathbb{N}$. That is

$g(x)= \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+ \frac{n}{n^2+n} \\ \leq \frac{n}{n^2}+\frac{n}{n^2}+...+\frac{n}{n^2} \\ = \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n} \\ = \frac{1}{n}(1+1+...+1) \\ =\frac{1}{n}(n) \\ =1$.

For the lower bound, we'll do essentially the samething, but for each $\frac{n}{n^2+k}$, we're going to use the estimate $\frac{n}{n^2+k} \geq \frac{n}{n^2+n}$, noting that there are $n$ of them, and then take a limit as $n \rightarrow \infty$.

Answer 3

You need to replace the given function by simpler ones that bracket it tightly and for which you can compute the limit. In the case of

$$\frac n{n^2+k}=\frac1{n+\dfrac kn},$$ as $1\le k\le n$, we have the inequalities

$$n\le n+\frac kn\le n+1,$$

leading to

$$n\frac1{n+1}\le\sum_{k=1}^n \frac n{n^2+k}\le n\frac1n.$$

Answer 4

Notice that $\frac{n}{n^2+1}\ge\frac{n}{n^2+2}\geq\frac{n}{n^2+3}\geq...\geq\frac{n}{n^2+n}$ - This is true because the numerators are equal, so the greater the denominator the smaller the number.

We have $n$ elements that are being summed, so from the aforementioned conclusion we can then conclude $n\times\frac{n}{n^2+1}=\frac{n^2}{n^2+1}\geq\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+...+\frac{n}{n^2+n}\geq n\times\frac{n}{n^2+n}=\frac{n^2}{n^2+n}$. So now we bounded given sum from above by $\frac{n^2}{n^2+1}$ and from below by $\frac{n^2}{n^2+n}$. Both boundaries go to $1$ as $n\to\infty$, and therefore from the squeeze theorem we can conclude that $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{n}{n^2+k}=1$.