How prove this $\frac{\sqrt{2}}{2}-\frac{1}{15}
Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$
  and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.

  
  
*
  
*Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$
  
*Find the value $F\left(\dfrac{\pi}{2}\right)$.
 A: Since the series $f$ converges normally on $\mathbb{R}$, we can integrate term-by-term. This gives
$$
F\left(\frac{\pi}{2}\right)=\sum_{n\geq 1}\frac{1}{\sqrt{n^3+n}}\int_0^{\pi/2}\cos (nx)dx=\sum_{k\geq 0}\frac{(-1)^k}{\sqrt{2}(2k+1)^\frac{3}{2}(2k^2+2k+1)^\frac{1}{2}}.
$$
The series converges absolutely to $S$. But this is an alternating series which satisfies Leibniz criterion (the relevant function is indeed decreasing on $(0,+\infty)$). So the sequence of partial sums $S_n$ alternates about $S$ and satisfies
$$
S_{2n+1}< S< S_{2n} \qquad \forall n\geq 0.
$$
In particular, we get
$$
\frac{\sqrt{2}}{2}-\frac{1}{15}<\frac{\sqrt{2}}{2}-\frac{1}{3\sqrt{5}\sqrt{6}}=S_1<S<S_0=\frac{\sqrt{2}}{2}.
$$
This answers 1.
I don't know if there is a closed form for $S$ and I am not alone.
A: I was delayed in posting this, so it is little more than a comment to julien's answer.
Since the series $\sum\limits_{n=1}^\infty\frac1{\sqrt{n^3+n}}$ is convergent, we can integrate term by term
$$
F(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n\sqrt{n^3+n}}
$$
$\sin((2k)\pi/2)=0$ and $\sin((2k+1)\pi/2)=(-1)^k$; therefore,
$$\begin{align}
F(\pi/2)
&=\frac1{\sqrt2}\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt{(2k+1)^3(2k^2+2k+1)}}\\
&=\frac1{\sqrt2}-\frac1{\sqrt{270}}+\dots
\end{align}
$$
By the alternating series test, the final sum is between $\frac1{\sqrt2}$ and $\frac1{\sqrt2}-\frac1{\sqrt{270}}$. That is,
$$
\frac1{\sqrt2}-\frac1{15}\lt\frac1{\sqrt2}-\frac1{\sqrt{270}}\lt F(\pi/2)\lt\frac1{\sqrt2}
$$
This evaluates to $0.6587329279592957$, but the ISC does not find a closed form.
