Let be the recursively defined sequence
where the first two terms are given and .
- Assume that this sequence converges and compute its limit in terms of the initial terms and .
- Prove that the sequences converges for every choice of and . Assume .
- Assume the limit exists, say
Then, we have
Hence, the value does not depend on , and we have
Now, taking the limit
- Proof. Assume and let . Now, we claim for that
We prove this claim by induction. For the case we have
So the statement is indeed true for the case . Now, assume the statement is true for some integer . Then we have
Therefore, the proposed formula indeed holds for all integers . We then express this formula as a sum,
So the terms of the sequence are the partial sums of this series. But, since is a constant (for any given and ) this is a geometric series; hence, it converges. Therefore, the terms of the sequence converge to a finite limit
You can prove b by an argument similar to the proof of Leibniz’s rule
Doesn’t the denominator in the sum in the last row need k-1 as exponent instead of n-1?
It is perhaps more clear to use induction directly to prove that an+2a(n+1)=a1+2a2 although you do so implicitly.