Sum of $\{n\sqrt{2}\}$

Solution 1:

Let $f$ be the function defined for a positive integer $N$ as follows: $$ f(N) = \sum_{1\le n\le N}\{n\sqrt 2\}\ . $$ We need a control of $f(N)-N/2$.

Here, i try to give arguments for the idea extracted from the following experimental computation:

Experiment: We approximate $\sqrt 2$ by using continued fractions, for such a fixed approximation, $P/Q$ say, we compute $f(Q)$ and $f(Q)-Q/2$. For the first few values of $Q$...

def f(N):
    return sum( [ RR(k*sqrt(2)).frac() for k in xrange(1, N+1) ] )

cf = continued_fraction( sqrt(2) )

# consider the first 15 "convergents" P/Q and compute f(Q) - Q/2 for them
for frac in cf.convergents()[:15]:
    P, Q = frac.numerator(), frac.denominator()
    print "sqrt(2) ~ %s :: Q = %s :: f(Q)-Q/2 ~ %s" % ( frac, Q, f(Q)-Q/2 )

Results:

sqrt(2) ~ 1 :: Q = 1 :: f(Q)-Q/2 ~ -0.0857864376269049
sqrt(2) ~ 3/2 :: Q = 2 :: f(Q)-Q/2 ~ 0.242640687119285
sqrt(2) ~ 7/5 :: Q = 5 :: f(Q)-Q/2 ~ -0.286796564403573
sqrt(2) ~ 17/12 :: Q = 12 :: f(Q)-Q/2 ~ 0.308657865101423
sqrt(2) ~ 41/29 :: Q = 29 :: f(Q)-Q/2 ~ -0.317100367703610
sqrt(2) ~ 99/70 :: Q = 70 :: f(Q)-Q/2 ~ 0.320702497141440
sqrt(2) ~ 239/169 :: Q = 169 :: f(Q)-Q/2 ~ -0.322176510488248
sqrt(2) ~ 577/408 :: Q = 408 :: f(Q)-Q/2 ~ 0.322790161566445
sqrt(2) ~ 1393/985 :: Q = 985 :: f(Q)-Q/2 ~ -0.323043813132131
sqrt(2) ~ 3363/2378 :: Q = 2378 :: f(Q)-Q/2 ~ 0.323148970494003
sqrt(2) ~ 8119/5741 :: Q = 5741 :: f(Q)-Q/2 ~ -0.323192510470562
sqrt(2) ~ 19601/13860 :: Q = 13860 :: f(Q)-Q/2 ~ 0.323210559656218
sqrt(2) ~ 47321/33461 :: Q = 33461 :: f(Q)-Q/2 ~ -0.323217967510573
sqrt(2) ~ 114243/80782 :: Q = 80782 :: f(Q)-Q/2 ~ 0.323221431812271
sqrt(2) ~ 275807/195025 :: Q = 195025 :: f(Q)-Q/2 ~ -0.323220559774200

So the difference is in these cases "surprisingly" in the interval $[-1,1]$.

The estimation: Let us try to convert these observation into a proof. Let $a$ be $\sqrt 2-1$, so $a\in(0,1/2)$. (I need $a<1$ below.) We have $\{n\sqrt 2\}=\{na\}$.

Let $P/Q$ be a rational approximation of $a$, an irreducible fraction, namely one provided by the continued fraction convergents of $a$, which form an "alternated sequence around $a$" $\dots P/Q, P'/Q', P''/Q'',\dots$, and the difference between two consecutive convergents $P/Q$ and $P'/Q'$ is $\pm 1/(QQ')$. In particular, $Q,Q'$ are relatively prime. We will also use in the following the "next convergent" $P'/Q'$.

Fix some natural $n$ with $0<n<Q$. So $nP/Q<n$ is not an integer. (If $Q|(nP)$, then $Q|n$.)

Assume now there is an integer between $na$ and $nP/Q$. Then the same integer is in the bigger interval between $nP'/Q'$ and $nP/Q$, which is of length $1/(QQ')<1/Q$. But from $nPQ\not\in\Bbb N$ to the next integer is a distance of at least $1/Q$. Contradiction. Our assumption is false.

So $na$ and $nP/Q$ have the same floor. Then $$ \left\{na\right\} = \left\{n\frac PQ+n\left(a-\frac PQ\right)\right\} $$ lies between the numbers $$ \left\{n\frac PQ\right\}\pm \underbrace{\left\{n\left(a-\frac PQ\right)\right\}}_{<n/(QQ')}\ . $$ Now let $n$ run in a set $S=S(Q)$ of $Q$ consecutive elements. Then $$ \begin{aligned} \sum_{\substack{n\in S(Q)\\Q\not|n}} \{na\} \text{ lies between } &\sum_{\substack{n\in S(Q)\\Q\not|n}} \left\{n\frac PQ\right\} \pm \sum_{\substack{n\in S(Q)\\Q\not|n}}\left\{n\left(a-\frac PQ\right)\right\} \\ \text{ thus between } &\sum_{0<n<Q} \frac nQ \pm \frac 1{QQ'}\sum_{n\in S(Q)}{\color{red}{n}} \\ = &\frac 12(Q-1) \pm \frac 1{QQ'}\sum_{n\in S(Q)}{\color{red}{n}} \ . \end{aligned} $$ (Later edit in red.) This can now be converted to a proof for the stated result as follows. I try to explain my feeling of a procedure using an example, $N=100000$ say.

Recall the following denominators "$Q$" of the convergents of $a=\sqrt 2-1$: $1$, $2$, $5$, $12$, $29$, $70$, $169$, $408$, $985$, $2378$, $5741$, $13860$, $33461$, $80782$, $195025$. Using $Q=80782$ and $S(Q)=(N-Q,N]\cap\Bbb Z$ we obtain a deviation from $\frac 12|S(Q)|$ which is less than (one plus) $\frac {N(N+1)/2}{QQ'}<\frac{Q'Q'/2}{QQ'}<2$. (Here, $Q'/Q\approx \sqrt 2+1$, which is the limit of the ration of two consecutive convergents.)

Then for the "new N", which is $N-Q=19281$ we use the "new Q" $13860$. And so on.

Using this we get a deviation of $f(N)$ from $N/2$ which is of the shape $2\log_{\sqrt 2+1} N$.

(Have to submit, hope that the idea is clear, this was more important for me than to write things rigurous, and possibly deflect from the idea.)

Solution 2:

Writing \begin{align} &\quad \sum_{n=1}^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) \\ &=\int_1^N \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right) {\rm d}n + {\cal O}(1) \tag{1} \\ &=\frac{1}{\sqrt{2}} \int_\sqrt{2}^{\sqrt{2}N} \left(x- \lfloor x \rfloor \right) {\rm d}x + {\cal O}(1) \\ &=\frac{1}{\sqrt{2}} \Bigg( N^2 - 1 - \Bigg[ \frac{\sqrt{2}N \left(\sqrt{2}N-1\right)}{2} + \frac{\left\{\sqrt{2}N\right\} \left(1-\left\{\sqrt{2}N\right\}\right)}{2} \\ &\quad - \frac{\sqrt{2} \left(\sqrt{2}-1\right)}{2} - \frac{\left\{\sqrt{2}\right\} \left(1-\left\{\sqrt{2}\right\}\right)}{2} \Bigg] \Bigg) + {\cal O}(1) \\ &=\frac{N}{2} + {\cal O}(1) \end{align} where we used $$ \int_0^x \lfloor t \rfloor \, {\rm d}t = \frac{x(x-1)}{2} + \frac{\left\{x\right\}\left(1-\left\{x\right\}\right)}{2} \, . $$ The order follows from the fact that $$ \frac{\left( \sqrt{2}N - \lfloor \sqrt{2}N \rfloor \right) + \left( \sqrt{2} - \lfloor \sqrt{2} \rfloor \right)}{2} = {\cal O}(1) \tag{2} $$ and $$\int_1^{N} \left( n\sqrt{2} - \lfloor n\sqrt{2} \rfloor \right)'' {\rm d}n \tag{3} \\ = \left( \sqrt{2}n - \lfloor \sqrt{2}n \rfloor \right)'\big|_{n=N} - \left( \sqrt{2}n - \lfloor \sqrt{2}n \rfloor \right)'\big|_{n=1} \\ =\sqrt{2} \sum_{k=-\infty}^{\infty}\left[ \delta\left( \sqrt{2} - k \right) - \delta\left( \sqrt{2}N - k \right) \right] $$ but this requires Euler-Maclaurin.

Due to the harsh critic about my somewhat heuristic argument I want to correct my approach as far as possible.

Set $$f(x)=x-\lfloor x \rfloor$$ and $$f_n(x)=\frac{1}{2} - \frac{1}{\pi} \sum_{k=1}^n \frac{\sin(2\pi k x)}{k} \, ,$$such that $$ \lim_{n\rightarrow \infty} f_n(x) = f(x) \, . $$ Since $f_n$ is differentiable we can use Euler-Maclaurin to calculate the sum $$ \sum_{k=1}^N f_n(ak) $$ with some $a$. The integral in (1) does not create much of an issue in the limit $n \rightarrow \infty$, since the limit is piecewise continuous and the integral can be splitted accordingly and then integrated. Also the limit of (2) is of ${\cal O}(1)$. So the problematic term which needs to be examined is the remainder $R_2$ ((3) was very heuristic) which can be written as $$ R_2=\int_1^N B_2\left(t-\lfloor t \rfloor\right) \frac{\rm d}{{\rm d}t} f_n'(at) \, {\rm d}t $$ neglecting unnecessary constants and $B_2$ is the second Bernoulli polynomial. We can express \begin{align} f_n'(at) &= 1-\sum_{k=-n}^{n} {\rm e}^{i2\pi k at} = 1 - \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \\ B_2\left(t-\lfloor t \rfloor\right) &= \left(t-\lfloor t \rfloor\right)(\left(t-\lfloor t \rfloor - 1\right) + \frac{1}{6} = \lim_{M \rightarrow \infty} \sum_{k=1}^M \frac{\cos(2\pi kt)}{\pi^2 k^2} \end{align} and integrate by parts $$ R_2=-B_2\left(t-\lfloor t \rfloor\right) \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \Bigg|_{1}^{N} - 2 \sum_{k=1}^{M} \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t \tag{4} \, . $$ For $a$ not an integer, the first term is bounded and ${\cal O}(1)$ as $n \rightarrow \infty$. The integral can be viewed as a functional for $n \rightarrow \infty$ in which case the Dirichlet kernel acts as a periodic delta-distribution $\sum_{m=-\infty}^{\infty} \delta(at-m)$ $$ \lim_{n \rightarrow \infty} \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t = \sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \frac{\sin\left(\frac{2\pi km}{a}\right)}{\pi k a} \, . $$

Evaluating $$ \sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \lim_{M\rightarrow\infty} -2\sum_{k=1}^{M} \frac{\sin\left(\frac{2\pi km}{a}\right)}{\pi k a} = \sum_{m=\lceil a \rceil}^{\lfloor Na \rfloor} \frac{2\{m/a\}-1}{a} \tag{5} $$

and using $\sum_{n=1}^{N}\{an\} = \frac{N}{2} + {\cal O}(?)$ this becomes ${\cal o}(N)$. So it is actually true ${\cal O}(1)$ does not follow.

We continue with the integral in (4) for $N$ integer \begin{align} &\quad -2\sum_{k=1}^\infty \int_1^N \frac{\sin(2\pi kt)}{\pi k} \frac{\sin\left((2n+1)\pi at\right)}{\sin\left(\pi at\right)} \, {\rm d}t \\ &=-4\sum_{m=1}^n \sum_{k=1}^\infty \int_1^N \frac{\sin(2\pi kt)\cos(2\pi m a t)}{\pi k} \, {\rm d}t \\ &=\frac{4}{\pi^2} \sum_{m=1}^n \sum_{k=1}^\infty \frac{\cos^2(N\pi m a)-\cos^2(\pi ma)}{k^2-m^2 a^2} \\ &=2\sum_{m=1}^n \left[ \frac{\cos^2(N\pi m a)-\cos^2(\pi ma)}{\pi^2 m^2 a^2} - \frac{\cot(\pi ma)\left(\cos^2(N\pi ma) - \cos^2(\pi ma)\right)}{\pi ma} \right] \end{align} where $a$ must be an irrational number now. The first term is ${\cal O}(1)$ for $n\rightarrow \infty$.

Any idea for the second?

It can be rewritten as \begin{align} &\quad \, \, \sum_{m=1}^n \frac{\cot(\pi ma)\left(\cos^2(N\pi m a) - \cos^2(\pi ma)\right)}{\pi ma} \\ &= \sum_{m=1}^n \frac{\cot(\pi ma)\left(\cos(N2\pi ma) - \cos(2\pi ma)\right)}{2\pi ma} \\ &= - \sum_{m=1}^{n}\cos(m\pi a) \, \frac{\sin\left((N+1)m\pi a\right)}{m\pi a} \, \frac{\sin\left((N-1)m\pi a\right)}{\sin(m\pi a)} \\ &= - \sum_{m=1}^{n} \frac{\sin\left((N+2)m\pi a\right)}{m\pi a} \, \frac{\sin\left((N-1)m\pi a\right)}{\sin(m\pi a)} + \sum_{m=1}^n \frac{ \sin\left(N2\pi ma\right) - \sin\left(2\pi ma\right) }{2\pi ma} \end{align} so the second sum is bounded again $\forall N$ and $n \rightarrow \infty$.

Not sure if it helps, but I have the following two identities for the sines $$ \frac{\sin\left((N-1)nx\right)}{\sin(nx)} = \sum_{l=2}^N \cos\left((N-l)nx\right) \cos^{l-2}(nx) $$ and $$ \frac{\sin\left((N-1)nx\right)}{\sin(nx)} = 1+2\sum_{l=1}^{\frac{N}{2}-1} \cos\left(l2nx\right) \, , $$ but evaluating this feels as if I'm running in circles.

I added a Figure of the RHS of (5) up to $N=10^6$ which doesn't look anything like $\log$, so either the numbers are just too small or I dont why it has to be $\log$.

Solution 3:

This is along the same line as another problem and my answer there: Determine whether $\sum_{n=1}^\infty \frac {(-1)^n|\sin(n)|}{n}$ converges

We need Koksma's inequality p. 143, Theorem 5.1 of 'Uniform Distribution of Sequences' by Kuipers and Niederreiter.

Theorem [Koksma]

Let $f$ be a function on $I=[0,1]$ of bounded variation $V(f)$, and suppose we are given $N$ points $x_1, \ldots , x_N$ in $I$ with discrepancy $$ D_N:=\sup_{0\leq a\leq b\leq 1} \left|\frac1N \#\{1\leq n\leq N: x_n \in (a,b) \} -(b-a)\right|. $$ Then $$ \left|\frac1N \sum_{n\leq N} f(x_n) - \int_I f(x)dx \right|\leq V(f)D_N. $$

To control the discrepancy, We apply the following theorem, for the sequence $x_n = n\alpha - \lfloor n \alpha \rfloor$. Note that with $\alpha = \sqrt 2$, it has a bounded partial quotients in its continued fraction.

Theorem 3.4 of p125 in the Kuipers and Niederreiter's book, states that

If an irrational real $\alpha$ has a bounded partial quotients, then the discrepancy $D_N$ satisfies $$ N D_N\ll \log N. $$

Then applying these to your problem with $f(x)=x$, we obtain $$ \bigg\vert\frac1N \sum_{n\leq N} \{ n\sqrt 2\} - \int_0^1 x \ dx\bigg\vert \ll \frac{\log N}N. $$

Therefore, we have an estimate of $$ \sum_{n\leq N} \{n\sqrt 2\}=\frac N2 + O(\log N). $$

To obtain a more precise estimate, we have $V(f)=1$ for $f(x)=x$. Also, Theorem 3.4 describes how $O(\log N)$ term behaves in a more precise way. Using those, we have $$ \Bigg\vert \sum_{n\leq N} \{n\sqrt 2\}-\frac N2 \Bigg\vert \leq 3+\left(\frac1{\log \xi}+\frac{2}{\log 3}\right)\log N. $$ Here we use $\xi=\frac{1+\sqrt 5}2$ and the continued fraction partial quotients are bounded by $2$ (It is in fact $[1;2,2,2,\ldots]$).