# Analysis series proof

1. Oct 11, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
Let (a_n) and (b_n) be sequences with bn > 0 for all n in N. Assume that lim (an/bn) = L with L > 0.
Prove that the series:
$$\sum{a_n}$$ converges if and only if $$\sum{b_n}$$ converges

2. Relevant equations

3. The attempt at a solution

Since this is an if and only if proof, i know that I need to first assume if the an series converges, then so does the bn series. Next, I need to show if I assume the second series converges, then so does the first.

Alright, this is what I tried so far:

Assume $$\sum{a_n}$$ converges.
Meaning that the sequence (an) converges to zero.

If (an) converges to zero that means that for epsilon greater than zero there exists an N in N so that for n>=N:

|an|<epsilon

Since we are assuming lim (an/bn) converges to L>0 that means for epsilon greater than zero, there exists an N in N so that for n>=N:
|(an/bn)-L|<epsilon

I think that to show this, I need to show that bn converges to zero as well, but I'm not too sure how to do that. Or also, there is a theorem stating if an and bn are sequences satisfying 0<=an<=bn for all n in N then
if $$\sum{b_n}$$ converges then $$\sum{a_n}$$ converges.

So I guess if I can show that an is larger than bn for all n in N then I can prove that the series converges as well.

Any help/hints would be great!!!

2. Oct 11, 2009

### n!kofeyn

Just because bn converges to zero, that does not mean the series of bn will converge. It looks like the theorem you stated might be a good approach, but I'm not for sure. Note that if you can show bn≤can, for some constant c, that might help. Because if an converges, then can converges as well. I haven't worked this proof out, but these are just some thoughts.

3. Oct 11, 2009

### dancergirlie

this was what I was thinking.... since the limit of an/bn>0 that means that an has to be >=bn for all n in N.
We're also assuming bn>0 for all n in N which would mean
0<=bn<=an for all n in N
Which would mean if we assume that the series of an converges, that implies that the series of bn would converge.

Is my assumption about the limits correct or am I way off?

Also, then how would I go about proving that an converges if bn does if an is greater than or equal to bn for all n in N. Would I have to use the fact that if bn converges to zero if the series of bn converges?

4. Oct 11, 2009

### dancergirlie

sorry i meant would i have to use the fact that if the series converges, then the sequence converges to zero

5. Oct 11, 2009

### Petek

Hi,

I'll give you some hints to prove that if $\sum{a_n}$ converges, then so does $\sum{b_n}$:

1. When proving that an infinite sum converges, you can disregard any finite number of terms (the value of the sum might be affected, but not the convergence).
2. If $\sum{a_n}$ converges and c is any constant, then $\sum{ca_n}$ also converges.
3. If $ca_n \geq 0$ for some constant c and if $|b_n| < ca_n$ for all but finitely many n, then $\sum{b_n}$ converges (Comparison Test).
4. We know that $\lim_{x\rightarrow \infty}\frac{a_n}{b_n} = L > 0$. Use the $\epsilon$ definition of convergence with $\epsilon = \frac{L}{2}$:

There exists an N > 0 such that if n > N, then

$$\left| \frac{a_n}{b_n} - L \right| < \frac{L}{2}$$

Simplify this inequality to conclude that

$$0 < b_n < \frac{a_n}{\frac{L}{2}}$$

Now use observation #3 above to conclude that $\sum{b_n}$ converges. Note that the simplified inequality implies that $a_n \geq 0$ for all but finitely many $a_n$, justifying the use of #3.

Let me know if this isn't clear, or if you would like more hints.

Petek

6. Oct 12, 2009

### dancergirlie

That's what confusing to me...
How do you get from
|an/bn-L|<L/2 to that inequality? I know that 0<bn, but how do yo justify that an/(L/2)>b?

Also, to show that an converges if bn converges would I have to find a constant c so that
0<c*seriesan< series of bn?

7. Oct 12, 2009

### dancergirlie

or actually i meant 0<can<bn for all n in N

8. Oct 12, 2009

### dancergirlie

oh wait nevermind, I understand, i was confused cause I was looking at the part showing that an<=(3/2)L*bn
But that is what I will use to prove that if bn converges then so does an. Thanks so much for your help!