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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.

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    $\begingroup$ Maybe I misunderstand but it's false for the first non-negative non-decreasing sequence I tried: 0, 0, 0, 0, ... $\endgroup$
    – Dan Piponi
    Commented May 31, 2018 at 22:53
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    $\begingroup$ thanks Dan Piponi, z_n should be a positive sequence. thank you for point this out. I have edited the post. Thank you very much. $\endgroup$
    – MathGuy
    Commented May 31, 2018 at 22:56
  • $\begingroup$ Somewhat similar sum (due Apery): $$\sum_{n=1}^\infty\frac{z_1z_2\cdots z_{n-1}}{(K+z_1)(K+z_2)\cdots(K+z_n)}=\frac1K.$$ $\endgroup$
    – Max Alekseyev
    Commented Jun 1, 2018 at 1:33

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The point is that the partial sum

$$ \sum_{n=1}^N \frac{z_n}{(K+z_1)\ldots(K+z_n)} K^n = K - \frac{K^{N+1}}{(K+z_1)\ldots(K+z_N)} $$

as is easy to prove by induction.

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  • $\begingroup$ How do you deduce the result then? $\endgroup$
    – lcv
    Commented Jun 1, 2018 at 8:35
  • $\begingroup$ @icv Bound the second term on the right by $K (1 + z_1/K)^{-N}$. $\endgroup$
    – Robert Israel
    Commented Jun 1, 2018 at 18:48
  • $\begingroup$ @icv use the fact that $z_n$ is non-decreasing. $\endgroup$
    – MathGuy
    Commented Jun 1, 2018 at 23:07

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