Does this sequence converges or diverges? This is the sequence:
$ \Sigma_{n=1}^\infty (\int_{0}^{\frac{1}{n}}\frac{sin^3t}{t}dt) $
I started by testing the Limit of the summand (hoping the result will be none 0) but I got:
$ \lim_{n \to \infty} a_n = \lim_{n \to \infty}  (\int_{0}^{\frac{1}{n}}\frac{sin^3t}{t}dt) => \sin x \le x \space where \space 0 \le x => \space \le \lim_{n \to \infty}  (\int_{0}^{\frac{1}{n}}t^2dt) = 0   $ 
so no luck here..
I'm only familiar with ratio, root and integral tests so how should I solve it?
Thanks 
 A: Hint:
$$
\left| \frac{\sin^3 t}{t}\right| \leq \frac{t^3}{t} = t^2,
$$
hence, if
$$
a_n := \int_0^{1/n} \frac{\sin^3 t}{t} \, dt
$$
you get the estimates
$$
0 \leq a_n \leq \int_0^{1/n} t^2 dt = \frac{1}{3n^3}\,.
$$
Now use the comparison theorem for series.
A: First, we calculate 
$$ \lim _{x\rightarrow 0} \frac{    \int _0^x \frac{\sin ^3t}{t} \text{d} t }{x^3}=^{\text{D.L} } = \lim _{x\rightarrow 0}\frac{\sin ^3x}{3x^3}=\frac{1}{3}.$$
Therefore, if we set $a_n=    \int _0^{\frac{1}{n}} \frac{\sin ^3t}{t} \text{d} t  $ and $b_n=\frac{1}{n^3}$ we get $$\lim_{n\rightarrow \infty} \frac{a_n}{b_n}=\frac{1}{3}$$
So, by the limit comparison test we can conclude that $a_n$ converges.
A: Twisted proof: as
$$x\longmapsto\int_0^{\frac1x}\frac{\sin^3t}t\,dt$$
is positive an decreasing,
$$\sum_{n=1}^\infty\int_{0}^{\frac1n}\frac{\sin^3t}t\,dt$$
is convergent iff
$$\int_1^\infty\int_0^{\frac1x}\frac{\sin^3t}t\,dt\,dx$$
is convergent. Using Fubini,
$$
\int_1^\infty\int_0^{\frac1x}\frac{\sin^3t}t\,dt\,dx =
\int_0^1\int_1^\frac1t\frac{\sin^3t}t\,dx\,dt =
\int_0^1\frac{\sin^3t}{t^2}\,dt\le\int_0^1 t\,dt = 1/2.
$$
