In a previous question Summing something with random and non-random parts I asked how one could estimate (specifically I am interested in an upper bound). $\sum_{k=1}^n e^{-k B_k^2} $ where $B_k$ is a uniformly distributed sequence on [0,1]. The response was to take the expected value of the summand. But what can I do if my $B_k^2$ are deterministic (and not random variables?). To clarify, by uniformly distributed I mean $\{B_k\}$ satisfies $ \lim_{n \rightarrow \infty} \frac{|[a,b] \cap \{B_1,B_2,\ldots B_n\}|}{n} = b-a$.
Edit: Suppose $B_n = \{ \alpha n \}$ is the fractional part of $\alpha n$ for some irrational number $\alpha$. By Weyl's Criterion I know that $\{B_n\}$ is uniformly distributed. From some numerical calculations it seems that there exists some $K > 0$ that only depends on $\alpha$ such that $\sum_{k=1}^n e^{-k B_k^2} < K \sqrt{n}$ for all sufficiently large $n$. How would I go about demonstrating this?
Edit 2 Here is an idea I had. $ \sum_{k=1}^n e^{-k B_k^2} = \sum_{k \leq n \: : e^{-k B_k^2} \leq n^{-1/2} } e^{-k B_k^2} + \sum_{k \leq n \: : e^{-k B_k^2} > n^{-1/2}} e^{-k B_k^2} $ which is less than $ \sum_{k = 1}^n n^{-1/2} + \sum_{k \leq n \: : e^{-k B_k^2} > n^{-1/2}} 1 $ Now the left sum is just $n^{1/2}$. Now $ \#\{e^{-k B_k^2} > n^{-1/2} : k \leq n\} = \#\{B_k < \sqrt{\log (n)/(2k)} : k \leq n\}$. I believe that $\#\{B_k < \sqrt{\log (n)/(2k)} : k \leq n\} \approx \int_1^n \sqrt{\frac{\log (n)}{2t}} \, dt$ Or maybe that integral should be a sum, I don't know. But that implies there exists $\epsilon > 0$ such that for sufficiently large $n$, $\#\{B_k < \sqrt{\log (n)/(2k)} : k \leq n\} < n^{1/2+\epsilon}$ So finally we have $ \sum_{k=1}^n e^{-k B_k^2} < n^{1/2} + n^{1/2+\epsilon} < 2 n^{1/2+\epsilon}$ Clearly the above argument has quite a few gaps in it. I am hoping that someone can give me some feedback as to whether or not the gaps can be filled.