Estimating a sum: Difference between revisions
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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> | :<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. | 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 | :<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}. | \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> | :<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 | 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> | ||
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Thus for instance if <math>\sigma = 0.7</math>, <math>t = 0.4</math>, and <math>N \geq N_0 = 2000</math>, one has | 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> | :<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 | One can compute numerically that the second term on the RHS is at most 0.0087, thus | ||
:<math> | :<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 | for all <math>N \geq 2000</math>. In particular | ||
:<math> | :<math>S_{0.7, 0.4}(N) \leq S_{0.7, 0.4}(2000) + 0.0087 \leq 1.706.</math> | ||
Similarly one has | Similarly one has | ||
:<math> | :<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 | The second term can be shown to be at most <math>0.253</math>, thus | ||
:<math> | :<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>. | for all <math>N \geq 2000</math>. | ||
Revision as of 23:12, 8 March 2018
For any [math]\displaystyle{ \sigma, t\gt 0 }[/math] and natural number [math]\displaystyle{ N }[/math], introduce the sum
- [math]\displaystyle{ 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]\displaystyle{ \sigma,t\gt 0 }[/math] and [math]\displaystyle{ N \geq N_0 \geq 1 }[/math]. Then
- [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ 1 \leq n \leq N }[/math], hence by the integral test one has
- [math]\displaystyle{ 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]\displaystyle{ a = e^u }[/math], the right-hand side becomes
- [math]\displaystyle{ \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]\displaystyle{ (1-\sigma) u + \frac{t}{4} (u^2 - 2u \log N) }[/math] is convex in [math]\displaystyle{ u }[/math], and is thus bounded by the maximum of its values at the endpoints [math]\displaystyle{ u = \log N_0, \log N }[/math]; thus
- [math]\displaystyle{ \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]\displaystyle{ \Box }[/math]
Thus for instance if [math]\displaystyle{ \sigma = 0.7 }[/math], [math]\displaystyle{ t = 0.4 }[/math], and [math]\displaystyle{ N \geq N_0 = 2000 }[/math], one has
- [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ N \geq 2000 }[/math]. In particular
- [math]\displaystyle{ S_{0.7, 0.4}(N) \leq S_{0.7, 0.4}(2000) + 0.0087 \leq 1.706. }[/math]
Similarly one has
- [math]\displaystyle{ 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]\displaystyle{ 0.253 }[/math], thus
- [math]\displaystyle{ S_{0.3, 0.4}(N) \leq S_{0.3, 0.4}(2000) + 0.253 \leq 3.469 }[/math]
for all [math]\displaystyle{ N \geq 2000 }[/math].
