# Homework Help: Convergence of a sequence.

1. Feb 6, 2012

### cragar

1. The problem statement, all variables and given/known data
Verify, using the definition of convergence of a sequence, that
the following sequences converge to the proposed limit.
a) $lim \frac{1}{6n^2+1}=0$
b) $lim \frac{3n+1}{2n+5}=\frac{3}{2}$
c) $lim \frac{2}{\sqrt{n+3}} = 0$
3. The attempt at a solution
A sequence $a_n$ converges to a real number a if for every ε there is
an N in the naturals such that whenever n≥N it follows that
$|a_n-a|< \epsilon$.
so for the first one I need $\frac{1}{6n^2+1}< \epsilon$
and then I turn it into $\frac{1}{\epsilon}<6n^2+1$
So i could pick an n large enough to make that happen.
on the second one I move the 3/2 over and then combine those
fractions with a common denominator and I get
$|\frac{-12}{4n+10}|< \epsilon$
Am I doing this right or am I way off.

2. Feb 6, 2012

### vela

Staff Emeritus
Good start, though the -12 should be -13. Now what?

3. Feb 6, 2012

### cragar

so I have $|\frac{-13}{4n+10}<\epsilon |$
for any epsilon I can pick an n large enough to make that true.

4. Feb 6, 2012

### vela

Staff Emeritus
You should explicitly find what N will work.

5. Feb 6, 2012

### cragar

that seems weird to me because $\epsilon$ could be anything
so how would any fixed N work.

6. Feb 6, 2012

### vela

Staff Emeritus
You're given some $\epsilon > 0$, and then you have to find some N for that given $\epsilon$ such that the implication holds, so N will generally depend on $\epsilon$.

$$\frac{1}{n} < \epsilon \qquad \Rightarrow \qquad n > \frac{1}{\epsilon},$$ you could choose N to be any integer greater than $1/\epsilon$.

7. Feb 6, 2012

### cragar

so then I just solve for n in terms of $\epsilon$
$.25(\frac{-13}{\epsilon }-10)<n$
thanks for your help by the way

8. Feb 6, 2012

### vela

Staff Emeritus
Since you're working with inequalities, you can do some simplifications to make the algebra less tedious:
$$\left|\frac{-13}{4n+10}\right| < \frac{16}{4n} = \frac{4}{n} < \epsilon$$So instead of that complicated expression you have, you can choose N to be an integer greater than 4/ε.