An interesting series converging to a constant

Question

Let $K>0$ be a constant. Suppose $\{z_n\}_{n=1}^\infty$ is a non-decreasing positive sequence. Then the series

$$\sum_{n=1}^\infty\frac{z_n}{(K+z_1)(K+z_2)\cdots(K+z_n)}K^n=K$$

This is a quite interesting result as the series is convergent and the limit doesn't depend on the choice of $\{z_n\}_{n=1}^\infty$, as long as it is non-decreasing and positive

I have run computer simulations and this result seems to hold. However, I am not sure how to prove it.

Answer 1

Consider $S_n - K$:

$$ -K + \dfrac{z_1}{z_1 + K}K + \dfrac{z_2}{(z_1 + K)(z_2 + K)} K^2 + ... + \dfrac{z_n}{(z_1 + K)(z_2 + K)...(z_n + K)}K^n = $$

$$-\dfrac{1}{z_1 + K}K^2 + \dfrac{z_2}{(z_1 + K)(z_2 + K)} K^2 + ... + \dfrac{z_n}{(z_1 + K)(z_2 + K)...(z_n+K)}K^n = $$

$$- \dfrac{1}{(z_1 + K)(z_2 + K)} K^3 + ... + \dfrac{z_n}{(z_1 + K)(z_2 + K)...(z_n+K)}K^n$$

and keep telescoping until you reach

$$S_n - K = -\dfrac{1}{(z_1 + K)(z_2 + K)...(z_n+K)}K^n$$

Edit

Sorry, I was interrupted and omit the most important part. This delta could be rewritten as $- \prod\dfrac{1}{1 + \frac{z_i}{K}}$, which vanishes as long as $\prod (1+ \dfrac{z_i}{K})$ diverges to infinity.

This is equivalent to $\sum z_i$ to diverge; you don't really need the non-decreasing sequence. As long as its sum diverges, your observation holds. Try $z_i = \dfrac{1}{i}$, for example.

Answer 2

The series can be transformed into a telescoping series in the following way:

$$ \sum_{n=1}^{\infty}\frac{z_n}{\prod_{j=1}^n(K+z_j)}K^n=\sum_{n=1}^{\infty}\left(\frac{K^n}{\prod_{j=1}^{n-1}(K+z_j)}-\frac{K^{n+1}}{\prod_{j=1}^n(K+z_j)}\right)=K $$