# find max,min,sup,inf..

#### evinda

##### Well-known member
MHB Site Helper
Hey!!! I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,....\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey!!! I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,....\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??
Hola!! Suppose we pick $m=n=2$, then I get $$\displaystyle B=\frac 2 4$$.
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?

#### evinda

##### Well-known member
MHB Site Helper
Hola!! Suppose we pick $m=n=2$, then I get $$\displaystyle B=\frac 2 4$$.
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?
Oh,yes!!!Right!!! So:
$$supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh,yes!!!Right!!! So:
$$supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that? Good!

... but suppose we pick $m=3, n=2$, then I get $$\displaystyle B=\frac 3 5$$.
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?

#### evinda

##### Well-known member
MHB Site Helper
Good!

... but suppose we pick $m=3, n=2$, then I get $$\displaystyle B=\frac 3 5$$.
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?
I understand..I think that we should pick then $n=2$..Right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I understand..I think that we should pick then $n=2$..Right?
Yep. Care to make a new guess about the largest value?
And what about the smallest value?

#### evinda

##### Well-known member
MHB Site Helper
Yep. Care to make a new guess about the largest value?
And what about the smallest value?
Hmm..I don't really know.. If we want the numerator to be as big as possible,we have to take $m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ .. Could you give me a hint how I can find it? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hmm..I don't really know.. If we want the numerator to be as big as possible,we have to take $m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ .. Could you give me a hint how I can find it? What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$

#### evinda

##### Well-known member
MHB Site Helper
What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$
It is equal to $1$!!! So,this is the supremum,right?? And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
It is equal to $1$!!! So,this is the supremum,right?? And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?
Yes! #### evinda

##### Well-known member
MHB Site Helper
Yes! And... there is no min and max of the set,right? #### Klaas van Aarsen

##### MHB Seeker
Staff member
And... there is no min and max of the set,right? Right!

MHB Site Helper

#### Klaas van Aarsen

##### MHB Seeker
Staff member
And how can I prove it?
Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$

#### evinda

##### Well-known member
MHB Site Helper
Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$
Great!!!Thank you very much!!! 