Is there any solution for the following recurrence problem

Question

Suppose that $(a_n)$ is an decreasing sequence of positive numbers. Is it possible to obtain a sequence $(b_n)$ of positive real numbers, s.t. $$b_{n}-b_{n+1}=n(a_{n}-a_{n+1}),$$
for all $n$ ? Or, is there any result that ensures the existence of such solution (if it is the case)?

Answer 1

Disregard this answer, I was incorrectly assuming the $a_n$ where positive integers.

Yes.

Set $b'_0:=0$, $b'_{n+1}:=b'_n - n(a_n-a_{n+1})$ for $n \ge 0$. So $\{b'_n\}$ is a sequence of numbers you are looking for, except that not all of its members are positive.

Since $\{a_n\}$ is a decreasing sequence of positive intergers, it must be finite. With $M:=\min(b'_n)$ we can finally define $b_n:=b'_n-M+1$ which is all positive and still fullfills the recurence relation.

Answer 2

This is not always possible, for example with $a_n=\frac{1}{n}$ you obtain: $$b_n-b_{n+1}=\frac{n}{n(n+1)}=\frac{1}{n+1}$$ so $\lim_{n \rightarrow + \infty}b_n=- \infty$.

As $(b_n)$ is decreasing it is positive iff its limit (in $\mathbb{R} \cup \{- \infty\}$) is positive.

Using an Abel transformation one can show that:

$$b_n-b_0=\sum_{k=0}^{n-1} (b_{k+1}-b_k)=(n-1) a_n -\sum_{k=1}^{n-1} a_k $$

so from the explicit expression of $(b_n)$ you can deduce conditions to have $(b_n)$ positive.

Answer 3

Summing, the left side telescopes to give $b_m-b_0 =\sum_{n=0}^{m-1} n(a_{n+1}-a_n) =\sum_{n=0}^{m-1} (( n+1)a_{n+1}-na_n-a_{n+1}) =ma_m-\sum_{n=0}^{m-1} a_{n+1} $.

What more do you want?