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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 a non-decreasing and positive sequence.

I have run computer simulations and this result seems to hold. However, I am not sure how to prove it. You may assume $\{z_n\}_{n=1}^\infty$ is also non-negative if that makes it easier to prove.

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

I have run computer simulations and this result seems to hold. However, I am not sure how to prove it. You may assume $\{z_n\}_{n=1}^\infty$ is also non-negative if that makes it easier to prove.

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 a non-decreasing and positive sequence.

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

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An interesting series converging to a constant

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

I have run computer simulations and this result seems to hold. However, I am not sure how to prove it. You may assume $\{z_n\}_{n=1}^\infty$ is also non-negative if that makes it easier to prove.