Understanding the squeeze theorem

[chwala]

TL;DR Summary

See attached

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? Of course i can see that for $n$ values greater than or equal to $4$ the theorem would be sufficient.

 

[fresh_42]

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 $n>3.$

 

[chwala]

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 .

 

[fresh_42]

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.

 

[chwala]

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.

 

[chwala]

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

 

[chwala]

* 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 $0$.

 

[chwala]

...on the highlighted i hope my reasoning is correct any further insight is welcome...refreshing a little bit...

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

$S = \dfrac{a}{1-r} = \dfrac {0.5}{1-0.5} = 1$

 

[chwala]

This series;

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.

 

[fresh_42]

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.

 

[fresh_42]

* 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?!?

 

[fresh_42]

...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.

 

[fresh_42]

Sorry, but this thread is getting a mess by discussing four posts in parallel.

Please open new threads. This one is closed.