February 17th, 2017, 01:56 AM  #1 
Newbie Joined: Jan 2017 From: India Posts: 12 Thanks: 1  Problem on limit
I have to find the value of limit x>infinity 1/[(2^n)*n!] So can I solve it this way? I mean we know that 1/2^n=1 where n tends to infinity and 1/n!=e1 where n tends to infinity (n starts from 1) Then limx>inf 1/[(2^n)*n!] = 1*(e1)=e1 Is this method correct? Or I am doing wrong? Last edited by skipjack; February 18th, 2017 at 02:55 AM. 
February 17th, 2017, 03:30 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,399 Thanks: 2105 Math Focus: Mainly analysis and algebra 
The method is valid, but your limit results are very wrong. Also, you are saying that $x$ goes to infinity but your expression is in $n$.

February 17th, 2017, 04:00 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,254 Thanks: 561  Quote:
Last edited by skipjack; February 18th, 2017 at 02:57 AM.  
February 17th, 2017, 11:01 AM  #4 
Newbie Joined: Jan 2016 From: Athens, OH Posts: 15 Thanks: 7 
I mean we know that 1/2^n=1 where n tends to inf and 1/n!=e1 where n tends to infinity (n starts from 1) From these statements, your problem is on series, not sequences. That is: $$\sum_{n=1}^{\infty}{1\over 2^n}=1\text{ and }\sum_{n=1}^{\infty}{1\over n!}=e1$$ I hope you are not confusing sequences and series. My advice is to learn a little Latex so you can post your questions intelligently. As to your proposed method: $$\sum_{n=1}^{\infty}a_n\sum_{n=1}^{\infty}b_n= \sum_{n=1}^{\infty}a_nb_n$$ for sequences $a_n$ and $b_n$. This doesn't work at all. Even for finite sums this is certainly false. Now for your problem: the Maclaurin series for $$e^{ax}=\sum_{n=0}^{\infty}{a^n\over n!}x^n$$ So for $x=1$, $$e^a1=\sum_{n=1}^{\infty}{a^n\over n!}$$ Thus for $a=1/2$, $$\sqrt{e}1=\sum_{n=1}^{\infty}{1\over 2^nn!}$$ 
February 17th, 2017, 08:50 PM  #5 
Newbie Joined: Jan 2017 From: India Posts: 12 Thanks: 1  
February 17th, 2017, 08:52 PM  #6 
Newbie Joined: Jan 2017 From: India Posts: 12 Thanks: 1  How? 1/2^n is an infinite geometric series. Their common ratio is 1/2 And its sum will be 1/2/(11/2)=1 Then how can it be 0? Last edited by skipjack; February 18th, 2017 at 03:09 AM. 
February 18th, 2017, 01:47 AM  #7 
Member Joined: Oct 2016 From: Melbourne Posts: 68 Thanks: 33 
What you have is NOT a series though! It's a SEQUENCE!


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