Thank you so much! This was so helpful. I knew how to determine if a limit diverges but I wasn't sure how to write a proof. This was exactly what I needed. Greetings from Chicago!
I know im late. Im doing this more for myself though. To prove the sequence a_n = 2^n diverges, let's first suppose it converges. Thus it is bounded. Let's say it's bounded for some M>0 Then, for all natural n, |a_n|
Thank you so much! This was so helpful. I knew how to determine if a limit diverges but I wasn't sure how to write a proof. This was exactly what I needed. Greetings from Chicago!
Awesome!
I appreciate you explaining your strategy after completing the proof. The hard part of Real Analysis is figuring out where to start!
Thank you! Analysis is a pain in the neck. But I say this every semester and I end up doing well just by being engaged in the struggle.
when u gotta surf at 2 but do ur homework at 3
Yeah!
I'm gonna go surf now !!!
@@TheMathSorcerer lmao saving my exam while surfing, u legend!!!
I just got back, I saw a SEAL in the water!! A huge head kept popping up staring at me, talk about freaking out hahahahahahs
Crazy experience❤️
This was very useful thanks, I was trying to prove that a sequence was not bounded supposing it was converging but too hard instead of this
it was super helpful thank u شكرا
You are welcome🔥
a very simple example to understand
شكرا
But can't M be a real positive number? Therefore M+1 is not necessarily an integer.
If M isn’t an integer, then there’s an integer that lies between M and M+1. For instance, if M=1.5, then between 1.5 and 2.5 is 2.
Can you solve this for me?➡️
is diverges to+infinite.
Plz😒
I know im late. Im doing this more for myself though.
To prove the sequence a_n = 2^n diverges, let's first suppose it converges. Thus it is bounded. Let's say it's bounded for some M>0
Then, for all natural n, |a_n|
@@saudinho1436 no man, instead of log_2 M^2, you should use log_2 M+1, cuz M^2>M is not always true.
and not the log itselft. log is not always a natural number, so you pick the smallest integer larger than it.
is it me or he looks like Jeff Bezos
♥️
:)
you look like jeff bezos
Why no epsilon? Just curious.
No need by using the fact that every sequence converging is bounded
Not convergence is not necessary condition for divergence