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Please advice how to prove that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. By this I mean there exists no positive real $B$ such that for any natural $n$ $$-B <\sum\limits_{i=1}^n \cos \sqrt{i} < B$$

UPD: it looks like the sum is not bounded (it follows from Euler–Maclaurin formula), so it seems it is just necessary to fill out details.

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    $\begingroup$ Are you sure that it's true? $\endgroup$ Oct 20, 2014 at 7:46
  • $\begingroup$ first idea:$$ \cos \sqrt{n} = 1 - {n \over 2} + {n ^2 \over 4!} -\dots $$ $\endgroup$
    – Blah
    Oct 20, 2014 at 7:50
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    $\begingroup$ I wouldn't be so sure that the sum is indeed bounded. Plotting seems to indicate oscillations that tend to get larger each time. $\endgroup$ Oct 20, 2014 at 7:57
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    $\begingroup$ $n$ in both places or $\sum\limits_{i=1}^n \cos \sqrt{i}$ ? $\endgroup$ Oct 20, 2014 at 7:59
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    $\begingroup$ @GerryMyerson: I just read your comment. I employed this very same idea in my answer. $\endgroup$
    – robjohn
    Oct 26, 2014 at 16:53

2 Answers 2

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Note that $$ (-1)^n\cos(x)\ge\frac12\quad\text{on}\quad\left[n\pi-\frac\pi3,n\pi+\frac\pi3\right] $$ Furthermore, $$ \left(n\pi+\frac\pi3\right)^2-\left(n\pi-\frac\pi3\right)^2=\frac{4\pi^2n}{3} $$ Thus, if we let $a_n=\left\lceil\left(n\pi-\frac\pi3\right)^2\right\rceil$ and $b_n=\left\lfloor\left(n\pi+\frac\pi3\right)^2\right\rfloor$, then $$ (-1)^n\sum_{k=a_n}^{b_n}\cos(\sqrt{k})\ge\frac{2\pi^2n}{3}-\frac12 $$ When a sequence changes by $\frac{2\pi^2n}{3}-\frac12$, up or down, either before or after, its absolute value must have been at least $\frac{\pi^2n}{3}-\frac14$.

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  • $\begingroup$ A nice elementary estimate! $\endgroup$ Oct 26, 2014 at 17:22
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Given any smooth function $f(x)$ defined over $(1 - \epsilon,\infty)$, we can rewrite its partial sum over $\mathbb{Z}_{+}$ as a Riemann-Stieltjes integral:

$$\sum_{k=1}^n f(k) = \int_{1^{-}}^{n^{+}} f(x) d\lfloor x \rfloor \tag{*1}$$

Let $\;B_n(x)\;$ be the $n^{th}$ Bernoulli polynomial and $\;P_n(x) = B_n(\{x\})\;$, we know

$$\lfloor x \rfloor = x - \{ x \} = x - \frac12 - B_1(\{x\}) = x - \frac12 - P_n(x)$$

and $P_n$ satisfies the relation $P'_n(x) = n P_{n-1}(x)$ for $n > 1$.

Using them, we can transform RHS of $(*1)$ by repeat integration by parts.

$$\begin{align} \sum_{k=1}^n f(k) = & \int_{1}^{n} f(x) dx - \int_{1^{-}}^{n+} f(x) dP_1(x)\\ = & \int_{1}^{n} f(x) dx - \bigg[ f(x) P_1(x) \bigg]_{1^{-}}^{n^{+}} + \frac12 \int_1^n f'(x) dP_2(x)\\ = & \int_{1}^n f(x) dx - \bigg[ f(x) P_1(x) - \frac12 f'(x) P_2(x) \bigg]_{1^{-}}^{n^{+}} - \frac{1}{3!} \int_1^n f''(x) dP_3(x)\\ \vdots &\\ = & \int_{1}^n f(x) dx + \underbrace{\frac12 (f(1) + f(n) ) + \sum_{k=1}^{p}\frac{B_{2k}}{(2k)!}\bigg[f^{(2k-1)}(x)\bigg]_1^n}_{C_{n,p}} + R_{n,p} \end{align} $$

The last line is the famous Euler-Maclaurin formula and the error term $R_{n,p}$ has the form: $$ R_{n,p} = -\frac{1}{(2p)!} \int_1^n f^{(2p)}(x) P_{2p}(x) dx = \frac{1}{(2p+1)!} \int_1^n f^{(2p+1)}(x) P_{2p+1}(x) dx \tag{*2} $$

For our problem, $f(x) = \cos\sqrt{x}$. We only need to keep the expansion up to $p = 1$. We have

$$\begin{align} \int_1^n f(x) dx &= 2\sqrt{n} \sin\sqrt{n} +2 \cos\sqrt{n}- 2(\sin 1 + \cos 1)\\ C_{n,1} &= \frac12(\cos\sqrt{n} + \cos 1) -\frac{1}{24}\left( \frac{\sin\sqrt{n}}{\sqrt{n}} - \sin 1\right) \end{align}$$

For the error term $R_{n,1}$, we will use the second form in $(*2)$.
Let $K = \sup\limits_{0 \le x \le 1}|P_3(x)| = \frac{1}{12\sqrt{3}}$, we have

$$\begin{align}|R_{n,1}| &= \frac{1}{3!}\left|\int_1^n \left( \frac{\sin\sqrt{x}}{8 x^{3/2}} + \frac{3\cos\sqrt{x}}{8 x^2} -\frac{3\sin\sqrt{x}}{8 x^{5/2}} \right) P_3(x) dx\right|\\ &\le \frac{K}{6}\int_1^n \left( \frac{1}{8 x^{3/2}} + \frac{3}{8 x^2} +\frac{3}{8 x^{5/2}} \right) dx\\ &\le \frac{K}{6}\int_1^\infty \left( \frac{1}{8 x^{3/2}} + \frac{3}{8 x^2} +\frac{3}{8 x^{5/2}} \right) dx\\ &= \frac{7K}{48} = \frac{7}{576\sqrt{3}} \approx 0.0070164 \end{align} $$ So for all $n$, we have $$\sum_{k=1}^n \cos\sqrt{k} = 2\sqrt{n} \sin\sqrt{n} + \frac{5}{2} \cos\sqrt{n}- \frac{47 \sin 1 + 36 \cos 1}{24} + \epsilon_n$$

with the error term $|\epsilon_n| \le \frac{1}{24\sqrt{n}} + |R_{n,1}| \le 0.05$.

From this, we see for large $n$, the behaviour of the partial sums $\sum\limits_{k=1}^n \cos\sqrt{k}$ are dominated by the term $2\sqrt{n}\sin\sqrt{n}$. It simply oscillate between $\pm 2\sqrt{n}$ as $n$ increases and hence unbounded.

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  • $\begingroup$ Brilliant, thanks a lot! $\endgroup$
    – Hedgehog
    Oct 26, 2014 at 16:09

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