Convergence of the series $ \sum\limits_n\int_{0}^{1/n}(\sinh(x))^{3/2}dx$ 
Does the series:$$ \sum_{n=1}^{\infty }\int_{0}^{1/n}(\sinh(x))^{3/2}dx$$ converge or diverge?

I intend to find $k$ such that $\sinh(x)\leq kx$ for all $x$ in $[0;\frac{1}{n}]$ but I can't find it. Could someone give me a hint to solve this problem?
 A: For $n\geq 2$, you have that $0\leq \sinh(x)^{3/2}\leq \sqrt x$.
Then,
$$\int_0^{1/n}\sinh(x)^{3/2}\mathrm d x\leq \int_0^{1/n}\sqrt{x}\mathrm d x=\frac23\frac{1}{n^{3/2}}.$$
Since $$\sum_{n=1}^\infty \frac{1}{n^{3/2}}<+\infty ,$$
the claim follow.
A: By the Cauchy-Schwarz inequality
$$ 0\leq\int_{0}^{\frac{1}{n}}\sinh(x)^{3/2}\,dx \leq \sqrt{\int_{0}^{\frac{1}{n}}x^{3/2}\,dx\int_{0}^{\frac{1}{n}}\left(\frac{\sinh(x)}{\sqrt{x}}\right)^3\,dx}\tag{1}$$
but over the interval $[0,1]$ the convexity of the $\sinh$ function grants $\sinh(x)\leq\sinh(1)x$, hence:
$$ \int_{0}^{\frac{1}{n}}\sinh(x)^{3/2}\,dx \leq \sinh(1)^{3/2}\int_{0}^{\frac{1}{n}}x^{3/2}\,dx = \frac{2\sinh(1)^{3/2}}{5}\cdot\frac{1}{n^{5/2}}\tag{2}$$
and:
$$ 0\leq \sum_{n\geq 1}\int_{0}^{\frac{1}{n}}\sinh(x)^{3/2}\,dx \leq \color{red}{\frac{2}{5}\,\zeta\left(\frac{5}{2}\right)\sinh(1)^{3/2}}\approx 0.68362.\tag{3}$$
An even tighter inequality can be deduced from 
$$\forall x\in(0,1),\qquad \sinh(x) < x\left[1+(\sinh(1)-1)\,x^2\right].\tag{4}$$
