Epislon Delta Proof: Limit of a Sequence

[b0it0i]

Homework Statement

prove: lim n-> inifity Sn = 0

Let Sn = (n+1)/(n^2 +1)

Homework Equations

(for all epsilon > o) (there exists N) (for all n) [ n>N => |Sn-0| < epsilon]

The Attempt at a Solution

i let epsilon be arbitrary, so we must show that there exists an N such that for all n [ n>N => |Sn-0| < epsilon

Find N

|Sn -0| = |(n+1)/(n^2 +1)| = (n+1)/(n^2 +1)

I'm completely stuck on this step. I'm not sure how to deal with inequalities where i can make it (some term / n) < epsilon

so I can't choose N= term/epsilon

Any help would be much appreciated

 

[NateTG]

Hint:
\frac{n+1}{n^2+1} &lt; \frac{n+1}{n^2-1}

 

[b0it0i]

with that in mind

do i say

n>2

so that 1. n^2 - 1 > 0
2. n+1 < n+n = 2n

n^2 >4
1/4 n^2> 1
-(1/4) n^2 < -1
n^2 -1> n^2 - (1/4) n^2 = (3/4) n^2

1/(n^2-1) < 4/(3n^2)

|Sn|< (2n)/(n^2-1) < (8n)/(6n^2)=4/(3n)

4/(3n) < epsilon
when
n> 4/(3 epsilon)

N = max {2, 4/(3 epsilon)}

 

[NateTG]

with that in mind

do i say
...
|Sn|< (2n)/(n^2-1) < (8n)/(6n^2)=4/(3n)

4/(3n) < epsilon
when
n> 4/(3 epsilon)

N = max {2, 4/(3 epsilon)}

What you have looks ok.

I was actually thinking:
S_n=\frac{n+1}{n^2+1}&lt;\frac{n+1}{n^2-1}=\frac{n+1}{(n+1)(n-1)}=\frac{1}{n-1}
Which leads to
N=\max(2,\frac{1}{\epsilon}+1)

Another good one is:
S_n=\frac{n+1}{n^2+1}\leq\frac{2n}{n^2}=\frac{2}{n}
(The inequality is clearly true since the numerator increases, and the denominator decreases, and both are positive.)
N=\max(1,\frac{2}{\epsilon})