# Difference between revisions of "Estimating a sum"

For any $\sigma, t\gt0$ and natural number $N$, introduce the sum

$F_{\sigma,t}(N) := \sum_{n=1}^N \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}}.$

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

Lemma 1 Let $\sigma,t\gt0$ and $N \geq N_0 \geq 1$. Then

$\sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} \leq F_{\sigma,t}(N) \leq \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} + \max( N_0^{1-\sigma - \frac{t}{4} \log \frac{N^2}{N_0}}, N^{1-\sigma - \frac{t}{4} \log N} ) \log \frac{N}{N_0}.$

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

$F_{\sigma,t}(N) \leq \sum_{n=1}^{N_0} \frac{1}{n^{\sigma + \frac{t}{4} \log\frac{N^2}{n}}} + \int_{N_0}^N \frac{1}{a^{\sigma + \frac{t}{4} \log \frac{N^2}{a}}}\ da.$

Making the change of variables $a = e^u$, the right-hand side becomes

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

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

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

The claim follows. $\Box$

Thus for instance if $\sigma = 0.7$, $t = 0.4$, and $N \geq N_0 = 2000$, one has

$F_{0.7, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} + \max( 2000^{0.3 - 0.1 \log \frac{N^2}{2000}}, N^{0.3 - 0.1 \log N}) \log \frac{N}{2000}.$

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

$F_{0.7, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.7 + 0.1 \log \frac{N^2}{n}}} + 0.0087$

for all $N \geq 2000$. In particular

$F_{0.7, 0.4}(N) \leq F_{0.7, 0.4}(2000) + 0.0087 \leq 1.706.$

Similarly one has

$F_{0.3, 0.4}(N) \leq \sum_{n=1}^{2000} \frac{1}{n^{0.3 + 0.1 \log \frac{N^2}{n}}} + \max( 2000^{0.7 - 0.1 \log \frac{N^2}{2000}}, N^{0.7 - 0.1 \log N}) \log \frac{N}{2000}.$

The second term can be shown to be at most $0.253$, thus

$F_{0.3, 0.4}(N) \leq F_{0.3, 0.4}(2000) + 0.253 \leq 3.469$

for all $N \geq 2000$.