# Thread: Use def of convergence

1. ## Use def of convergence

Use the def of convergence to prove the sequence converges.

Def: A sequence $\{a_n\}_{n\in\mathbb{N}}$ converges to an $A\in\mathbb{R}$ iif for each $\epsilon >0$ there exists $N\in\mathbb{N} \ \text{such that} \ \forall n\geq N \ \text{we have} \ |a_n-A|<\epsilon$

$\left\{5+\frac{1}{n}}\right\}_{n=1}^{\infty}$

So A = 5. Now, using the def.

Let $\epsilon >0$ and there is N such that for $n\geq N, \ \left | 5+\frac{1}{n}-5\right |=\left |\frac{1}{n}\right |<\epsilon$

$\Rightarrow \frac{1}{n}\leq\frac{1}{N}<\epsilon$

$\Rightarrow N>\frac{1}{\epsilon}$

Is that it?

2. ## Re: Use def of convergence

Originally Posted by dwsmith
Use the def of convergence to prove the sequence converges.

Def: A sequence $\{a_n\}_{n\in\mathbb{N}}$ converges to an $A\in\mathbb{R}$ iif for each $\epsilon >0$ there exists $N\in\mathbb{N} \ \text{such that} \ \forall n\geq N \ \text{we have} \ |a_n-A|<\epsilon$

$\left\{5+\frac{1}{n}}\right\}_{n=1}^{\infty}$

So A = 5. Now, using the def.

Let $\epsilon >0$ and there is N such that for $n\geq N, \ \left | 5+\frac{1}{n}-5\right |=\left |\frac{1}{n}\right |<\epsilon$

$\Rightarrow \frac{1}{n}\leq\frac{1}{N}<\epsilon$

$\Rightarrow N>\frac{1}{\epsilon}$

Is that it?
Instead: We need to find such N.

And... N is natural, but 1/epsilos is not always... so you can write N>[1/epsilon]+1 where [x] is the smallest integer than not greater than x.

3. ## Re: Use def of convergence

The definition says one has to be able to find an $N$ given an $\epsilon$. You showed that one can take any $N>1/\epsilon$, e.g., $1/\epsilon$ rounded up.

You only need to change the direction of the arrows.

$N > 1/\epsilon$ and $n > N \Rightarrow |a_n-A|=\frac{1}{n}\leq\frac{1}{N}<\epsilon$

4. ## Re: Use def of convergence

How would I handle:

$\{2^{-n}\}$

Would I say:

$N>\ln\left(\frac{1}{\epsilon}-2\right)\mbox{?}$

5. ## Re: Use def of convergence

$2^{-n}=\frac{1}{2^n}$

And prove by induction : $2^n>n$ for $n>2$

$2^n>n$ or $\frac{1}{2^n}<\frac{1}{n}$

6. ## Re: Use def of convergence

Originally Posted by Also sprach Zarathustra
2^{-n)=1/(2^n)
I am aware of that.

7. ## Re: Use def of convergence

It should be any $N>\ln\left(\frac{1}{\epsilon}\right)/\ln2=-\frac{\ln\epsilon}{\ln2}=-\log_2\epsilon$.

I am always trailing Zarathustra, but...

You can also take any $N>\frac{1}{\epsilon}$ because $2^n>n\Rightarrow 2^{-n}<1/n$, so when $n>N>\frac{1}{\epsilon}$, we have $2^{-n}<1/n<1/N<\epsilon$.

8. ## Re: Use def of convergence

one doesn't have to find "an optimal N" (although finding such an N can be evidence of a certain...cleverness), just one that's "big enough".

in Zarathustra's example, the sequence {1/n} bounds {2^(-n)} from above, since {1/n} goes to 0 in the limit,

and both sequences are positive, we have a special case of the "squeeze theorem" at work: {0} < {2^(-n)} < {1/n}.

this is useful to remember when proving convergence: if the sequence is bounded by another sequence that makes the ε-algebra

easier, go ahead and find an N for the "bigger sequence".

of course the derivation of an optimal N is the case of 2^(-n) isn't that hard:

we want: |2^(-n)| = 2^(-n) < ε. so log2(2^(-n)) < log2(ε) (log base 2 is increasing so preserves order)

so -n < log2(ε), so we want n > -log2(ε) (the alternate form in emakarov's post no doubt comes from re-writing:

2^(-n) = e^(-nln(2)), which has the advantage of being in the form of the standard log function, which makes it more

transparent that ln is increasing, after all, it's derivative is 1/x, which is > 0 for all x in the domain of ln).