Hi Haitao,

thanks, and yes, f is a sequence of reals.  I'm following a similar path
(proof by contradiction), but I don't understand the last step: for any
m, the partial sum of f goes up by 1 at n > m, why f must be summable? I
think for every monotonic sequence such properties hold.


Il 01/03/19 17:24, Haitao Zhang ha scritto:
> You say f is a function. From the context I assume the domain is Nat or
> f is a sequence. Mathematically speaking, you should form the partial
> sums of f ( sum f from 1 to n), which is a monotonic sequence of nats.
> Now proof by contradiction: if your conclusion doesn’t hold, for any m,
> you can find n > m, such that f n = 1. Then the partial sum goes up by 1
> at n. As m is arbitrary, f is not summable.
> However I don’t know what is the best way to carry out the above proof
> in hol as I am not familiar with the relevant libraries yet.
> Haitao
> On Friday, March 1, 2019, Chun Tian (binghe) <binghe.l...@gmail.com
> <mailto:binghe.l...@gmail.com>> wrote:
>     Hi,
>     I'm blocked at the following goal:
>     I have a function f returning either 0 or 1.  Now I know the infinite
>     sum of f is finite, i.e.
>             suminf f < PosInf       (or `summable f` speaking reals)
>     How can I prove the set of {x | f x = 1} is finite, or after certain
>     index m all the rest f(n) are zeros?
>             ∃m. ∀n. m ≤ n ⇒ (f n = 0)
>     If I use CCONTR_TAC (proof by contradiction), I can easily derive the
>     following 2 assumptions using results I established in my previous
>     similar questions:
>             INFINITE N
>             ∀n. n ∈ N ⇒ (f n = 1)
>     but still I've no idea how to derive a contradiction with `suminf f <
>     PosInf` by proving `suminf f = PosInf`...
>     Thanks,
>     Chun Tian

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