Proof that $\sum\limits_{i=1}^n \cos \sqrt{i}$ is unbounded. 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. 
 A: 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 $\frac{\pi^2n}{3}-\frac14$.
A: 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.
