Understanding the squeeze theorem

  • I
  • Thread starter chwala
  • Start date
  • Tags
    Theorem
In summary: S = \dfrac{a}{1-r} = \dfrac {0.5}{1-0.5} = 1##Same problem with L'Hôpital as before. See the first one.* i do not want to post a new thread as the questions are related...Also for##c_n = \dfrac{ \ln (n^2)}{n}##here i used the L 'Hopital's rule i.e##\lim_{x→∞} \dfrac {\ln
  • #1
chwala
Gold Member
2,648
351
TL;DR Summary
See attached
1666871362473.png


The theorem is pretty clear...out of curiosity i would like to ask...what if we took ##n-3## factors...then the theorem would not be true because we shall have;

##[n!=6⋅4⋅5⋅6 ... n]##

and

##[2^n = 8⋅2⋅2⋅2 ...2]##

What i am trying to ask is at what point do we determine the number of terms to consider? Is there a general rule when using Squeeze Theorem or is it a matter of one using common sense? :wink: Of course i can see that for ##n## values greater than or equal to ##4## the theorem would be sufficient.
 
Physics news on Phys.org
  • #2
chwala said:
TL;DR Summary: See attached

What i am trying to ask is at what point do we determine the number of terms to consider?
We have ##\dfrac{-1}{2^n}\leq \dfrac{(-1)^n}{n!}\leq \dfrac{1}{2^n}.## In order to squeeze something, we need ##\dfrac{1}{n!}<\dfrac{1}{2^n}## which is equivalent to ##2^n<n!## and this isn't true for ##n\leq 3,## hence ##n>3.##
 
  • Like
Likes malawi_glenn and chwala
  • #3
fresh_42 said:
We have ##\dfrac{-1}{2^n}\leq \dfrac{(-1)^n}{n!}\leq \dfrac{1}{2^n}.## In order to squeeze something, we need ##\dfrac{1}{n!}<\dfrac{1}{2^n}## which is equivalent to ##2^n<n!## and this isn't true for ##n\leq 3,## hence ##n>3.##
That is exactly what I am saying, ...I agree... then it's just a matter of logical thinking ( checking the comparison and so forth) ...what about sequences that may have alternating terms? would the theorem still apply? I may need to explore further. Thanks @fresh_42 .
 
  • #4
chwala said:
That is exactly what I am saying, ...I agree... then it's just a matter of logical thinking ( checking the comparison and so forth) ...what about sequences that may have alternating terms? would the theorem still apply? I may need to explore further. Thanks @fresh_42 .
Be careful. If ##a_n\leq b_n \leq c_n## - alternating or not - and ##L:=\lim_{n \to \infty}a_n## and ##M:=\lim_{n \to \infty}c_n## then ##L=M## implies ##\lim_{n \to \infty}b_n=L. ## The important condition is ##L=M,## not convergence. If ##L\neq M## then we cannot make any conclusions without further information like e.g. monotony.
 
  • Like
Likes malawi_glenn and chwala
  • #5
fresh_42 said:
Be careful. If ##a_n\leq b_n \leq c_n## - alternating or not - and ##L:=\lim_{n \to \infty}a_n## and ##M:=\lim_{n \to \infty}c_n## then ##L=M## implies ##\lim_{n \to \infty}b_n=L. ## The important condition is ##L=M,## not convergence. If ##L\neq M## then we cannot make any conclusions without further information like e.g. monotony.
Nice, i see that it does not matter whether a sequence is alternating or not as the Absolute Value theorem will apply. Cheers mate.
 
  • #6
Still on this, instead of using a similar sequence...can't we just use limits straightaway? i am asking this in reference to another example;

For this case; we may have

##\lim_{n→∞} \left[(-1)^n \dfrac {1}{n!}\right] =0##

My next example is on determining the divergence or convergence of the sequence;

##b_n = (-1)^n \dfrac{n}{n+1}##

it follows that;

##\lim_{n→∞} \left[(-1)^n \dfrac {n}{n+1}\right] =\left[\lim_{n→∞} (-1)^n\right]⋅\left[\lim_{n→∞} \dfrac {n}{n+1}\right]=∞⋅1=∞##

thus sequence diverges.

i hope i am doing it right...cheers
 
Last edited:
  • #7
* i do not want to post a new thread as the questions are related...

Also for

##c_n = \dfrac{ \ln (n^2)}{n}##

here i used the L 'Hopital's rule i.e

##\lim_{x→∞} \dfrac {\ln (x^2)}{x}= \lim_{x→∞} \left[\dfrac{2}{x}\right]=0##

The sequence converges to ##0##.
 
Last edited:
  • #8
...on the highlighted i hope my reasoning is correct any further insight is welcome...refreshing a little bit...

1667044331567.png


On the first highlighted part to arrive there, they made use of;

##Sn= \dfrac{a(1-r^n)}{1-r}= \dfrac{0.5(1-0.5^n)}{1-0.5}=1- \dfrac{1^n}{2^n}= \dfrac{2^n -1 }{2^n}##

on the second highlighted they made use of L'Hopital's rule;

##\lim_{x→∞} \dfrac {2^x-1}{2^x}= \lim_{x→∞} \left[\dfrac{2^x \ln 2}{2^x \ln 2}\right]=1##

Another approach;

Also ...the series is geometric, therefore ##r=\dfrac{1}{2}## because ##0< |r|< 1##, the series converges and its sum is

##S = \dfrac{a}{1-r} = \dfrac {0.5}{1-0.5} = 1##
 
Last edited:
  • #9
This series;

1667045494871.png
can also be expressed as one fraction i.e

##S_n= \dfrac{1}{n(n+1)}## ...the partial sums converge to ##1##. We have ##[0.5, 0.666, 0.75, 0.8, ...)##

##\lim_{n→∞} \dfrac{1}{n(n+1)}=\lim_{n→∞}\left[ \dfrac{A}{n} + \dfrac{B}{1+n}\right] ##

With

##A=1##

##A+B=0, ⇒B=-1##

then the results given will be realized...this is just but a different way of me looking at the problem...of course the text approach is straightforward no dispute there.
 
  • #10
chwala said:
Still on this, instead of using a similar sequence...can't we just use limits straightaway? i am asking this in reference to another example;

For this case; we may have

##\lim_{n→∞} \left[(-1)^n \dfrac {1}{n!}\right] =0##

My next example is on determining the divergence or convergence of the sequence;

##b_n = (-1)^n \dfrac{n}{n+1}##

it follows that;

##\lim_{n→∞} \left[(-1)^n \dfrac {n}{n+1}\right] =\left[\lim_{n→∞} (-1)^n\right]⋅\left[\lim_{n→∞} \dfrac {n}{n+1}\right]=∞⋅1=∞##

thus sequence diverges.

i hope i am doing it right...cheers
You cannot write it that way. ##\lim_{n \to \infty}(-1)^n## does not exists, and even less is it ##\infty .## To show divergence, you could e.g. simply show that ##|a_n-a_{n-1}|>0.5.## Limit formulas do not help in such a case.
 
  • #11
chwala said:
* i do not want to post a new thread as the questions are related...

Also for

##c_n = \dfrac{ \ln (n^2)}{n}##

here i used the L 'Hopital's rule i.e

##\lim_{x→∞} \dfrac {\ln (x^2)}{x}= \lim_{x→∞} \left[\dfrac{2}{x}\right]=0##

The sequence converges to ##0##.
You need to justify why you can transition from discrete to continuous. Also, how exactly did you apply L'Hôpital? Wikipedia's version is
$$
\lim_{x \to x_0}\dfrac{f'(x)}{g'(x)}=c \Longrightarrow \lim_{x \to x_0}\dfrac{f(x)}{g(x)}=c
$$
Yours seem to be different?!?
 
  • #12
chwala said:
...on the highlighted i hope my reasoning is correct any further insight is welcome...refreshing a little bit...

View attachment 316322

On the first highlighted part to arrive there, they made use of;

##Sn= \dfrac{a(1-r^n)}{1-r}= \dfrac{0.5(1-0.5^n)}{1-0.5}=1- \dfrac{1^n}{2^n}= \dfrac{2^n -1 }{2^n}##

on the second highlighted they made use of L'Hopital's rule;

##\lim_{x→∞} \dfrac {2^x-1}{2^x}= \lim_{x→∞} \left[\dfrac{2^x \ln 2}{2^x \ln 2}\right]=1##

Another approach;

Also ...the series is geometric, therefore ##r=\dfrac{1}{2}## because ##0< |r|< 1##, the series converges and its sum is

##S = \dfrac{a}{1-r} = \dfrac {0.5}{1-0.5} = 1##
Same problem with L'Hôpital as before. See above.
 
  • #13
Sorry, but this thread is getting a mess by discussing four posts in parallel.

Please open new threads. This one is closed.
 
  • Like
Likes SammyS

1. What is the squeeze theorem?

The squeeze theorem, also known as the sandwich theorem or the pinching theorem, is a mathematical theorem used to prove the limit of a function. It states that if two other functions, one above and one below the function in question, have the same limit at a certain point, then the function in question must also have that same limit at that point.

2. How is the squeeze theorem used?

The squeeze theorem is used to evaluate limits of functions that are difficult to evaluate directly. By finding two functions that "squeeze" the function in question, we can determine its limit by looking at the limits of those two functions.

3. What are the conditions for using the squeeze theorem?

In order to use the squeeze theorem, the function in question must be sandwiched between two other functions, and all three functions must have the same limit at a certain point. Additionally, the two "squeezing" functions must be continuous at that point.

4. Can the squeeze theorem be used to prove the limit of a function does not exist?

Yes, the squeeze theorem can also be used to prove that the limit of a function does not exist. If the two "squeezing" functions have different limits at a certain point, then the function in question must also not have a limit at that point.

5. Are there any limitations to using the squeeze theorem?

Yes, the squeeze theorem can only be used to evaluate limits at a single point. It cannot be used to evaluate limits as x approaches infinity or negative infinity. Additionally, it can only be used for functions that are sandwiched between two other functions, and the two "squeezing" functions must be continuous at the point in question.

Similar threads

Replies
15
Views
2K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
3
Views
1K
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
226
Replies
2
Views
2K
Replies
15
Views
1K
Back
Top