My Math Forum Convergence of a sequence

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 March 9th, 2013, 04:47 AM #1 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Convergence of a sequence How to prove n/(2 n+sqrt(n)) converges.
 March 9th, 2013, 04:49 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Convergence of a sequence Hint : Use L'hospital to see if the limit exists at infinity.
 March 9th, 2013, 04:55 AM #3 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Convergence of a sequence How do I prove this using the epsilon delta definition of limits?
March 9th, 2013, 04:58 AM   #4
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Re: Convergence of a sequence

Quote:
 Originally Posted by martemac How do I prove this using the epsilon delta definition of limits?
You want epsilon-delta? Then it's not worth being in the Real analysis section.

 March 9th, 2013, 05:00 AM #5 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Convergence of a sequence I am sorry. I am new to the forum. Where should it be?
March 9th, 2013, 05:09 AM   #6
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Re: Convergence of a sequence

Quote:
 Originally Posted by martemac I am sorry. I am new to the forum. Where should it be?
It should be in the calculus forum, but I think one of our moderators will move it, you don't have to worry about it.

 March 9th, 2013, 05:10 AM #7 Newbie   Joined: Mar 2013 Posts: 6 Thanks: 0 Re: Convergence of a sequence Thank you.
March 9th, 2013, 09:18 AM   #8
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Re: Convergence of a sequence

Quote:
 Originally Posted by mathbalarka It should be in the calculus forum, but I think one of our moderators will move it, you don't have to worry about it.
it is a real analysis question , I am currently reading a real analysis book , which deals with convergence of sequences using that definition ...

$\lim_{n\to \infty}\frac{n}{2n+\sqrt{n}}=\frac{1}{2}$

so we must prove that :

$\|\frac{n}{2n+\sqrt{n}}-\frac{1}{2}\|<\epsilon$

$\|\frac{\sqrt{n}}{2(2n+\sqrt{n})}\|<\epsilon$

$\|\frac{\sqrt{n}}{2(2n+\sqrt{n})}\|\leq\|\frac{\sq rt{n}}{2(2n)}\|=\|\frac{1}{4\sqrt{n}}\|$

we can easily prove that :

$\|\frac{1}{4\sqrt{n}}\|<\epsilon$

Hence the sequence :

$\frac{n}{2n+\sqrt{n}}$ converges ...

March 9th, 2013, 09:27 AM   #9
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Re: Convergence of a sequence

Quote:
 Originally Posted by zaidalyafey ... it is a real analysis question...
Yes, that's why I decided to leave the topic here, and the OP has duplicated the topic in the Calculus forum as well. Normally we discourage duplicate posts, but I made an exception in this case.

March 9th, 2013, 09:29 AM   #10
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Re: Convergence of a sequence

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by martemac I am sorry. I am new to the forum. Where should it be?
It should be in the calculus forum, but I think one of our moderators will move it, you don't have to worry about it.
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I think it is in the right section, as far as I know, sequences with the ?-? definition are not taught in high schools. In Greece for example, students in high school, learn about the definition briefly, meaning the they will never use this definition in the final exams and on the other hand only the definion of limits of functions (not sequences, although they are much alike).

Speculating from mathbalarka's posts, such things are taught in his school from the elementary grades.

Zaidalyafey gave the answer with a better $n_0$, but here is another way.

$a_{n}=\frac{n}{2n+\sqrt{n}}=\frac{\frac{n}{n}}{\fr ac{2n}{n}+\frac{\sqrt{n}}{n}}=\frac{1}{2+\frac{\sq rt{n}}{n}}$

We will prove that $\frac{\sqrt{n}}{n}\to 0$,

for every ?>0, we will find a suitable $n_{0}=n(\epsilon)$ such that $\left|\frac{\sqrt{n}}{n}\right|<\epsilon$. From the last inequality we get $\left|\frac{\sqrt{n}}{n}\right|<\epsilon\Rightarro w \left|\frac{1}{\sqrt{n}}\right|<\epsilon\Rightarro w n>\frac{1}{\epsilon^2}$. So we proved that for all ?>0 and $n>\frac{1}{\epsilon^2}$, $|a_{n}-\frac{1}{2}|<\epsilon$, meaning that $a_{n}\to\frac{1}{2}$.

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