Difference between revisions of "Estimating a sum"

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For any <math>\sigma, t>0</math> and natural number <math>N</math>, introduce the sum
 
For any <math>\sigma, t>0</math> and natural number <math>N</math>, introduce the sum
  
:<math>F_{\sigma,t}(N) := \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}.</math>
+
:<math>F_{\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.   
 
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>0</math> and <math>N \geq N_0 \geq 1</math>.  Then
 
'''Lemma 1'''  Let <math>\sigma,t>0</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 F_{\sigma,t}(N) \leq
+
:<math> \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} \leq F_{\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}.
+
\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>
 
</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
 
'''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>F_{\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>
+
:<math>F_{\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
 
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>
+
:<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
 
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>
 
:<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>

Revision as of 15:46, 4 March 2018

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

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

Similarly one has

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

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