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Suppose $f: \mathbb{N} \longrightarrow \mathbb{N}$ is a non-decreasing function such that $\sum_{t=1}^{\infty} 2^{-f(t)} < \infty$. Is it possible to find another non-decreasing function $g: \mathbb{N} \longrightarrow \mathbb{N}$, such that $g \longrightarrow \infty$ and $\sum_{t=1}^{\infty} g(t)2^{-f(t)} < \infty$? My idea : let $g$ be a sufficiently slow growing function (floor(log log log($f(t)$)) for example?) However, I don't have a proof that my idea works.

Note: I edited it after seeing the comments

Tejas Bhojraj
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  • Just make $g(t)$ increase from 1 to 2. – Michael May 05 '19 at 15:22
  • @Michael $g$ has to be on the naturals – Tejas Bhojraj May 05 '19 at 15:23
  • @Don. Do you mean that there is an $f$ for which there is no increasing $g$ which works? – Tejas Bhojraj May 05 '19 at 15:27
  • @Quantum My apologies, I misread the question. Ignore my comment. – Rushabh Mehta May 05 '19 at 15:30
  • @Quantum You can actually always find such an $g$. All you need to do is find a subsequence of $t$ such that $f(t_i)<2*f(t_{i+1})$. Then, you can define $g$ according to these indices pretty easily. – Rushabh Mehta May 05 '19 at 15:33
  • @Don. thanks; I will try it out! – Tejas Bhojraj May 05 '19 at 15:34
  • Well, if you consider $g(t)$ is $O(\min(t^k,f(t)))$, for some $k$ and $f$ be strictly increasing, then you have what you need by $t-$th root test. Indeed, since $f$ is an increasing function of integers, we have that $f(t)\geq t$. Also $$\sqrt[t]{\frac{g(t)}{2^{f(t)}}}\leq\frac{\sqrt[t]{g(t)}}{2}\to\frac{1}{2}.$$ – Vassilis Markos May 05 '19 at 15:36
  • @Quantum : To me, $g$ on the naturals means $g$ is defined on the naturals, so take $g(n) = 2-1/n$ for $n \in {1, 2, 3, ...}$. Indeed that increases from 1 to 2. Else, if you want $g(n) \in \mathbb{N}$, I don't know how you are assuming $\log\log(n)$ is a natural number. – Michael May 05 '19 at 15:47
  • @Michael. Sorry, I meant $g$ is from naturals to naturals. And I meant take the ceiling or floor. Will modify my question now – Tejas Bhojraj May 05 '19 at 16:13
  • @DonThousand $f(t_{i})<2*f(t_{i+1})$ will hold for all $i$ since $f(t_{i})/2 < f(t_{i}) \leq f(t_{i+1})$. Did you mean something else? – Tejas Bhojraj May 05 '19 at 16:36
  • I meant it the other way around, oops. Flip the $i$ and $i+1$. – Rushabh Mehta May 05 '19 at 16:38
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    See https://math.stackexchange.com/a/452074/120267 – Kitegi May 05 '19 at 17:46

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