Inductive Sequences---plz help!

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ku1005

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Define the sequence {an}1 to infinity inductively by:

a1=0

a(n+1) = 1/(n+1) + an/2 , for every n>=1

(i) Proove by induction that 2/(n+2) <=an+1 < = an , for evry n>=3


(ii) Deduce from part (i) that the sequence {an}1 to infinty converges to 0 and and the series

(sum to infinity) of an diverges.


If anyone could help here that would be great!!...thanks for your time!
 
I am not sure what is meant by:

defining a sequence inductively".

Please showus some work so that we can venture a guess.
 
This is good problem. What have done on it?
Show us where you have problems.
At least start the induction.
 
well....this question really stumped me, since I havt seen anything this hard before.

Usually when we do these questions we come up with a sequence which satisfies for all n, starting from and including n =1, then we proove via induction (ie proove that s(1), and then assume s(k) is true, then show that s(k+1) is true) so that the sequence we came up with is true for all k.


I treid looking at the first few terms of the sequence:

a1= 0 = 0

a2 = 1/2 =1/2

a3 = 1/3 + 1/4 =7/12

a4 = 1/4 + 1/6 + 1/8 =13/24

a5 = 1/5 + 1/8 + 1/12 + 1/16 = 113/240

but i cant see anything jumping out at me

an = 1/n + .....??? the truth is i not really not sure, once a sequence is found it makes life simpler but, are you able to spot a sequence which satisfies all n??
 
ku1005 said:
Define the sequence {an}1 to infinity inductively by:

a1=0

a(n+1) = 1/(n+1) + an/2 , for every n>=1

(i) Proove by induction that 2/(n+2) <=an+1 < = an , for evry n>=3
Click to expand...
If you write the definition of a(n+1) in the form \(\displaystyle a_{n+1} \;= \;\frac12\left(\frac2{n+1}\; +\; a_n\right),\) then you notice that a(n+1) is the arithmetic mean of a(n) and 2/(n+1). Since we want a(n+1) to be less than a(n), this suggests that a(n) ought to be greater than 2/(n+1).

So, let's try to prove by induction that

...............\(\displaystyle a_n\;\geq \;\frac2{n+1}\) for all \(\displaystyle n \geq3\). ..... (*)

The base case n=3 is okay, since 7/12 > 1/2. For the inductive step,

...............\(\displaystyle a_{n+1}\; = \;\frac1{n+1}\; + \;\frac{a_n}2 \;\geq \;\frac1{n+1} \;+\; \frac1{n+1}\; = \;\frac2{n+1}\; > \;\frac2{n+2}.\)

That completes the inductive proof of (*) . From the first paragraph above, that tells us that a(n+1) <= a(n). Also, if you replace n by n+1 in (*) then you see that 2/(n+2) <= a(n+1). So we have proved the two inequalities in (i).
 
thanks heaps for your help, took me a while to understand how to approach it, but you make it seem easy, thanks again!
 
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