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 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!=e-1 where n tends to infinity (n starts from 1) Then limx>inf 1/[(2^n)*n!] = 1*(e-1)=e-1 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: 6,702 Thanks: 2177 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
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Quote:
 Originally Posted by Calculus96 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
No, we don't. 1/2^n goes to 0 as n goes to infinity.

Quote:
 Originally Posted by Calculus96 and 1/n!=e-1 where n tends to infinity (n starts from 1) Then limx>inf 1/[(2^n)*n!] = 1*(e-1)=e-1 Is this method correct? Or I am doing wrong?

Last edited by skipjack; February 18th, 2017 at 02:57 AM.

 February 17th, 2017, 11:01 AM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 43 Thanks: 25 I mean we know that 1/2^n=1 where n tends to inf and 1/n!=e-1 where n tends to infinity (n starts from 1) From these statements, your problem is on series, not sequences. That is: $$\sum_{n=1}^{\infty}{1\over 2^n}=1\text{ and }\sum_{n=1}^{\infty}{1\over n!}=e-1$$ 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 $x=1$, $$e^a-1=\sum_{n=1}^{\infty}{a^n\over n!}$$ Thus for $a=1/2$, $$\sqrt{e}-1=\sum_{n=1}^{\infty}{1\over 2^nn!}$$ Thanks from greg1313
February 17th, 2017, 08:50 PM   #5
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Quote:
 Originally Posted by v8archie 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 $n$.
That's a mistake!

February 17th, 2017, 08:52 PM   #6
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Quote:
 Originally Posted by Country Boy No, we don't. 1/2^n goes to 0 as n goes to infinity.
How?
1/2^n is an infinite geometric series. Their common ratio is 1/2
And its sum will be
1/2/(1-1/2)=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: 77 Thanks: 35 What you have is NOT a series though! It's a SEQUENCE!

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