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.