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I hope this question is appropriate for MathOverflow, if not please remove it.

I have the following problem which is taken from the paper "Sharp Conditional Bound for Moments of the Riemann Zeta Function" by Adam J. Harper (pdf version).

Main Question

Let $\beta_i=\frac{20^{i-1}}{(\log\log T)^2}$ for all $i\geq 1$, $k\geq 1$ and $\mathcal{I}=1+\max\{i\mid \beta_i\leq e^{-1000k}\}$, then $$ \sum_{j=1}^{\mathcal{I}-1}e^{-0.01k/\beta_j}\ll_k 1. $$

How I tried to approach the problem

I tried to estimate this sum using a modified version of a standard way of tackling sums (for example see equation 1.62 in the book 'Analytic Number Theory' by Iwaniec and Kowalski) in number theory.

Lemma

Let $f\in\mathcal{C}^{1}(\mathbb{R})$ monotonic increasing, $f\geq 0$ and $\{a_i\}_{i=1}^{N}\subset \mathbb{R}$ an increasing sequence of points such that $\min\{a_{j+1}-a_j\}\geq 1$. Then $$ \sum_{j=1}^{N}f(a_j)\leq\int_{a_1}^{a_N+1}f(t)\,dt. $$

Proof. By monotonicity of the function we have that for $1\leq j \leq N$ $$ f(a_j)\leq \int_{a_j}^{a_j+1}f(t)\,dt. $$ Since by hypothesis we have $\min\{a_{j+1}-a_j\}\geq 1$, it follows that $a_j+1\leq a_{j+1}$ and therefore $$ f(a_j)\leq \int_{a_j}^{a_j+1}f(t)\,dt\leq \int_{a_j}^{a_{j+1}}f(t)\,dt, \qquad\qquad \forall 1\leq j\leq N-1, $$ hence \begin{equation*} \begin{split} \sum_{j=1}^{N}f(a_j)&=\sum_{j=1}^{N-1}f(a_j) + f(a_N)\leq\sum_{j=1}^{N-1}\int_{a_j}^{a_{j+1}}f(t)\,dt+\int_{a_N}^{a_N+1}f(t)\,dt=\int_{a_1}^{a_N+1}f(t)\,dt \end{split} \end{equation*} proving our claim.


We can now use the above lemma to estimate our sum under consideration. We have first to prove that the hypothesis of the lemma are verified. We take $f(x)=e^x$ which is clearly monotonic increasing, positive and $\mathcal{C}^1(\mathbb{R})$. We then take $a_j=-0.01k/\beta_j$. This sequence is monotonically increasing since also $\{\beta_i\}$ is an increasing sequence. Moreover, using that $\beta_{j+1}=20\beta_j$, we have \begin{split} a_{j+1}-a_j&=-\frac{0.01k}{\beta_{j+1}}+\frac{0.01k}{\beta_j}\\ &=0.01k\left(-\frac{1}{\beta_{j+1}}+\frac{1}{\beta_j}\right)\\ &=0.01k\left(-\frac{1}{20\beta_j}+\frac{1}{\beta_j}\right)\\ &=0.01k\left(\frac{19}{20\beta_j}\right)\\ &\geq 0.01k\left(\frac{19}{20\beta_{\mathcal{I}-1}}\right)\\ &\geq 0.01k\frac{19}{20} e^{1000k}\\ &\geq \frac{0.019}{20}e^{1000}\\ &\geq 1 \end{split} where we have used that $\beta_j\leq \beta_{\mathcal{I}-1}$ for all $j\leq \mathcal{I}-1$ and that $\beta_{\mathcal{I}-1}\leq e^{-1000k}$ which implies that $\frac{1}{\beta_{\mathcal{I}}}\geq e^{1000k}$. We can therefore apply the lemma and we get \begin{split} \sum_{j=1}^{\mathcal{I}-1}e^{-0.01k/\beta_j}&\leq \int_{-0.01k/\beta_1}^{-0.01k/\beta_{\mathcal{I}-1}+1} e^x\,dx\\ &\leq \int_{-\infty}^1 e^x\,dx\\ &=e \end{split} where we have used that $-0.01k/\beta_j\leq 0$ for all $j$. This concludes the proof of our main question.

Does it make sense or there is some mistake that I cannot spot? My doubt is: do I need really $\beta_{\mathcal{I}−1}≤e^{−1000k}$ to prove it? I think I do not, at least this is what my proof seems to suggest if there are no mistakes. Thanks for your help!

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    $\begingroup$ From $\beta_{\mathcal{I}-1}\leq e^{-1000k}$ it does not follow that $\frac{1}{\beta_{\mathcal{I}}}\geq e^{1000k}$. It only follows that $\frac{1}{\beta_{\mathcal{I}-1}}\geq e^{1000k}$, which is sufficient, however. This is because $j\leq\mathcal{I}-1$, hence in bounding $a_{j+1}-a_j$ one can use $\frac{19}{20\beta_{\mathcal{I}-1}}$ in place of $\frac{19}{20\beta_{\mathcal{I}}}$. Note that $\beta_{\mathcal{I}-1}\leq e^{-1000k}$ was used in this argument. $\endgroup$
    – GH from MO
    Commented Aug 29, 2019 at 5:10
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    $\begingroup$ Well, one does not need $\beta_{\mathcal{I}-1}\leq e^{-1000k}$ for this argument, it suffices that $\beta_{\mathcal{I}-1}\leq 10^{-3}$, say. However, Harper needs that $\log(1/\beta_\mathcal{I})\geq 900k$ for deducing his (3.1). You are summing the right hand side of Harper's (3.1), but one also needs the validity of (3.1). $\endgroup$
    – GH from MO
    Commented Aug 29, 2019 at 5:26
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    $\begingroup$ Sorry for the typo. You are right, to get something negative in the exponent multiplying $1/\beta_j$ we need to have$$\left(2k-\frac{\log(1/\beta_{j+1})}{420}\right)\leq 0$$ which implies that $\beta_{j+1}\leq e^{-840k}$. $\endgroup$
    – asd
    Commented Aug 29, 2019 at 7:35

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