Limit of recursive series

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sambellamy

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I have a problem that asks to find the value of the limit of an. I am given:

a1=4 and an+1= (1/4)(an+1)

I am not sure how to take the limit. I think i disregard a1 and just take the limit of an+1. I started to spell out all of the terms but could not find a suitable pattern. I found that I have 4n-1 as the denominator, but not sure what to do with the numerator. I have:

a2= (1/4)(4 + 1) = 1 + 1/4

a3= (1/4)(1 + 1/4 + 1) = (1/4)(1/4 + 2) = 1/16 + 1/2 = (5+4)/42

a4= (1/4)(1/16 + 1/2 + 1) = 1/64 + 1/8 + 1/4 = 25/64

a5= (1/4)(1/64 + 1/8 + 1/4 + 1) = 1/256 + 1/32 + 1/16 + 1/4 = 89/256

I can see a relationship that is messed up by the second-to-last term. in my mind it should be 1/4n-2.

Please help! Am I going about this in the right way?
 
Think this way: IF there is a limit, A, then taking the limit of both sides of the recursion equation \(\displaystyle \lim_{n\to\infty} a_{n+1}= \lim_{n\to\infty}(1/4)(a_n+ 1)\). Those "\(\displaystyle \{a_{n+1}\}\)" and "\(\displaystyle \{a_n\}\)" are the same sequence so have the same limit. \(\displaystyle A= (1/4)(A+ 1)= 1/4+ A/4\). Solve that for A.

Of course, that was assuming that there was a limit- that the sequence does converge. If you can prove, by induction, say, that this sequence is decreasing, since it is clearly bounded below by 0, it must converge.
 
I see now that if an+1 = (1/4)(an+1), then

limn->∞ an+1 = limn->∞ (1/4)(an+1). Are you saying that solving for limn->∞
an will give the same limit?
Is that where A = (1/4)(A + 1) came from? it seems like you are using the same variable for an and an+1.

Anyway I solved the capital-A equation and got A = 1/3. I deduce that this is the limit of the sequence, is that correct?
I have mostly been dealing with series - just to make sure, the limit of the sequence is the nth value (an), correct?
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Another way is to work backwards
an+1 = \(\displaystyle \frac{1}{4}(a_n + 1) = \frac{1}{4}a_n + \frac{1}{4}\)
= \(\displaystyle \frac{1}{4^2}(a_{n-1} + 1) + \frac{1}{4}=\frac{1}{4^2}a_{n-1} + \frac{1}{4^2} + \frac{1}{4}\)
= \(\displaystyle \frac{1}{4^3}(a_{n-2} + 1) + \frac{1}{4^2} + \frac{1}{4} = \frac{1}{4^3}a_{n-2} + \frac{1}{4^3} + \frac{1}{4^2} + \frac{1}{4}\)
= ...
= \(\displaystyle \frac{1}{4^j}(a_{n-j+1} + 1) + \frac{1}{4^{j-1}} + ... + \frac{1}{4} = \frac{1}{4^j}a_{n-j+1} + \frac{1}{4^j} + \frac{1}{4^{j-1}} + ... + \frac{1}{4}\)
= ...
= \(\displaystyle \frac{1}{4^n}(a_{1} + 1) + \frac{1}{4^{n-1}} + ... + \frac{1}{4} = \frac{1}{4^n}a_{1} + \frac{1}{4^n} + \frac{1}{4^{n-1}} + ... + \frac{1}{4}\)
--> (\(\displaystyle 1 + \frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + ... \)) - 1
 
HallsOfIvy, I need some help. What am I solving for when solving for A? Also, when I try to solve limn->∞ (1/4)(an+1), i just get (1/4)limn->∞(an) + 1/4. This does not help at all - I am back to an. What am I missing?
 
sambellamy said:
HallsOfIvy, I need some help. What am I solving for when solving for A? Also, when I try to solve limn->∞ (1/4)(an+1), i just get (1/4)limn->∞(an) + 1/4. This does not help at all - I am back to an. What am I missing?
Click to expand...
When you are solving for A, you are solving for the limit that you want to find! I said "IF there is a limit, A, Because "\(\displaystyle a_n\)" and "\(\displaystyle a_{n+1}\)" give the same limit: \(\displaystyle \lim_{n\to\infty} a_{n+1}= \lim_{n\to\infty} a_n\) which I have called "A".
 
Aha! I now see that:

limn->∞ an = limn->∞ an+1 = L (the letter I am accustomed to)

limn->∞ an+1 = limn->∞ (1/4)(an+1)

= (1/4) limn->∞ an + limn->∞ (1/4)

= (1/4) L + (1/4) = L

and then we get L = 1/3

Thank you for your help, and your patience!
 
Notice that, as I said before, we are assuming the sequence does converge. If, in fact, the series does not converge then that answer is non-sense. We still need to prove that series does converge. You should be able to prove, using proof by induction, that the sequence is decreasing. Since it is obviously bounded below by 0, it must converge.
 
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