Estimating a sum

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For any [math]\sigma, t\gt0[/math] and natural number [math]N[/math], introduce the sum

[math]S_{\sigma,t}(N) := \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}}.[/math]

This sum appears a number of times in the Polymath15 project, and it is therefore of interest to estimate it.

Lemma 1 Let [math]\sigma,t\gt0[/math] and [math]N \geq N_0 \geq 1[/math]. Then

[math] \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} \leq S_{\sigma,t}(N) \leq \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} + \max( N_0^{1-\sigma - \frac{t}{4} \log \frac{N^2}{N_0}}, N^{1-\sigma - \frac{t}{4} \log N} ) \log \frac{N}{N_0}. [/math]

Proof The left-hand inequality is obvious. To prove the right-hand inequality, observe (from writing the summand as [math]\frac{\exp( \frac{t}{4} (\log N - \log n)^2}{n^\sigma \exp(\frac{t}{4} \log^2 N)}[/math]) that the summand is decreasing for [math]1 \leq n \leq N[/math], hence by the integral test one has

[math]S_{\sigma,t}(N) \leq \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} + \int_{N_0}^N \frac{1}{a^{\sigma + \frac{t}{4} \log \frac{N^2}{a}}}\ da.[/math]

Making the change of variables [math]a = e^u[/math], the right-hand side becomes

[math]\sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} + \int_{\log N_0}^{\log N} \exp( (1-\sigma) u + \frac{t}{4} (u^2 - 2u \log N) )\ du.[/math]

The expression [math](1-\sigma) u + \frac{t}{4} (u^2 - 2u \log N)[/math] is convex in [math]u[/math], and is thus bounded by the maximum of its values at the endpoints [math]u = \log N_0, \log N[/math]; thus

[math]\exp( (1-\sigma) u + \frac{t}{4} (u^2 - 2u \log N) ) \leq \max( N_0^{1-\sigma - \frac{t}{4} \log \frac{N^2}{N_0}}, N^{1-\sigma - \frac{t}{4} \log N} ).[/math]

The claim follows. [math]\Box[/math]

Thus for instance if [math]\sigma = 0.7[/math], [math]t = 0.4[/math], and [math]N \geq N_0 = 2000[/math], one has

[math]S_{0.7, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} + \max( 2000^{0.3 - 0.1 \log \frac{N^2}{2000}}, N^{0.3 - 0.1 \log N}) \log \frac{N}{2000}. [/math]

One can compute numerically that the second term on the RHS is at most 0.0087, thus

[math]S_{0.7, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} + 0.0087[/math]

for all [math]N \geq 2000[/math]. In particular

[math]S_{0.7, 0.4}(N) \leq S_{0.7, 0.4}(2000) + 0.0087 \leq 1.706.[/math]

Similarly one has

[math]S_{0.3, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.3 + 0.1 \log \frac{N^2}{n}}} + \max( 2000^{0.7 - 0.1 \log \frac{N^2}{2000}}, N^{0.7 - 0.1 \log N}) \log \frac{N}{2000}.[/math]

The second term can be shown to be at most [math]0.253[/math], thus

[math]S_{0.3, 0.4}(N) \leq S_{0.3, 0.4}(2000) + 0.253 \leq 3.469[/math]

for all [math]N \geq 2000[/math].