Help with this sequence problem please

[bobopedic]

lim n
n->infinity -------- = 1/2
2n+1

Using the definition of the limit of a sequence prove the above

Anyone have any idea?
Thanks in advance, guys.

 

[tkhunny]

Pick a positive number, often called \(\displaystyle \delta\).

Demonstrate that for some value of n, the quantity n/(2n+1) is closer to 1/2 than this \(\displaystyle \delta\).

Are you then done?

 

[Deleted member 4993]

lim n
n->infinity -------- = 1/2
2n+1


\(\displaystyle \lim_{n \to \infty}\frac{n}{2n+1}\)

Using the definition of the limit of a sequence prove the above

Anyone have any idea?
Thanks in advance, guys.

Try dividing the numerator and the denominator by 'n'

 

[bobopedic]

Pick a positive number, often called \(\displaystyle \delta\).

Demonstrate that for some value of n, the quantity n/(2n+1) is closer to 1/2 than this \(\displaystyle \delta\).

Are you then done?

This is what I did, I think I'm wrong, though.

Prove lim n-->infinity of n/(2n+1) =1/2

Given E>0, choose N=???

| n/(2n+1) -1/2 < E <-----> | 2n-2n-1/2(2n+1) | < E

Then 1/(4n+2) < E <-----> 4n+2 > 1/E

Then 4n < 1/E - 2 (=1-2E/E)

Then I get n > 1-2E/4E

So...given E < 0, Choose N= 1-2E/4E

Is that correct or am I way off?
And if I'm write I understand how I got there but how does this prove the sequence of the limit?

 

[bobopedic]

lim n
n->infinity -------- = 1/2
2n+1


\(\displaystyle \lim_{n \to \infty}\frac{n}{2n+1}\)

Using the definition of the limit of a sequence prove the above

Anyone have any idea?
Thanks in advance, guys.

Try dividing the numerator and the denominator by 'n'

I know that

You get 1/(2 + 1/n) and as n approaches infinity 1/n approaches 0 making it 1/2, but I think I have to use the definition of the limit.

 

[bobopedic]

Hmmm I think I might get it, I think I pick a random E that's bigger than 0, say two. Now N=1-2E/4E which is now 1-4/8 which is 3/8. Now n > 3/8, so like 1, 2, 3. Now if I plug 1,2,3 into the original n/2n+1, I'll get 1/3, 1/5, 1/7.....so if E >0, and n>N, then the original is less than < E?

 

[bobopedic]

Anyone?

 

[tkhunny]

Hmmm I think I might get it, I think I pick a random E that's bigger than 0, say two. Now N=1-2E/4E which is now 1-4/8 which is 3/8. Now n > 3/8, so like 1, 2, 3. Now if I plug 1,2,3 into the original n/2n+1, I'll get 1/3, 1/5, 1/7.....so if E >0, and n>N, then the original is less than < E?

This is very much the idea. Good work.

Technical Point: You can't pick "random" numbers. Just make it small. The implication is that you could have picked something smaller - extending to arbitrarily small.

 

[bobopedic]

Hmmm I think I might get it, I think I pick a random E that's bigger than 0, say two. Now N=1-2E/4E which is now 1-4/8 which is 3/8. Now n > 3/8, so like 1, 2, 3. Now if I plug 1,2,3 into the original n/2n+1, I'll get 1/3, 1/5, 1/7.....so if E >0, and n>N, then the original is less than < E?

This is very much the idea. Good work.

Technical Point: You can't pick "random" numbers. Just make it small. The implication is that you could have picked something smaller - extending to arbitrarily small.

Thanks. Was my answer to the question correct, though? Any mistakes?

And also in the proof is it n > N or n greater or equal to N?

 

[tkhunny]

Anyone?

Have a little patience. We are just volunteers, here. We don't pay anyone to sit and watch the board. We just wander by when we can.

 

[tkhunny]

It needs to be ALL n greater than N. Once you get closer than delta, it must STAY closer than delta.