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February 17th, 2017, 01:56 AM   #1
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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.
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February 17th, 2017, 03:30 AM   #2
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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 $n$.
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February 17th, 2017, 04:00 AM   #3
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Quote:
Originally Posted by Calculus96 View Post
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 View Post
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.
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February 17th, 2017, 11:01 AM   #4
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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 $a=1/2$,
$$\sqrt{e}-1=\sum_{n=1}^{\infty}{1\over 2^nn!}$$
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February 17th, 2017, 08:50 PM   #5
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Quote:
Originally Posted by v8archie View Post
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!
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February 17th, 2017, 08:52 PM   #6
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Quote:
Originally Posted by Country Boy View Post
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.
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February 18th, 2017, 01:47 AM   #7
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What you have is NOT a series though! It's a SEQUENCE!
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