Here the coefficients $a_n$ are positive and decreasing in a neighborhood of $+\infty$. Moreover, there is the limit
$$\lim_{n\rightarrow\infty } \frac{a_n }{n^{-3/2}}= \alpha > 0 .$$
What is the asymptotic behavior of $f$ as $t\rightarrow 0^+$? I guess it is in the form
$$f\simeq c_0 - c_1 x^{1/2}, $$
where $c_0 = \sum_n a_n =f (0)$.
The $x^{1/2}$ behavior is conjectured by comparison with the series
$$ \sum_n \frac{\cos n x }{ n^{3/2}} . $$
For such kind of series, we can use the trick of Mellin transform to extract its asymptotic behaviors, as done in this thread
asymptotic behavior of the series $f(t) = \sum_{n\geq 1 } \cos(n t )/\sqrt{n}$
and we get the $x^{1/2}$ behavior.
I was thinking of proving the conjecture in this way. First separate out the constant term, and consider the series
$$g (x) = -\sum_n a_n (\cos nx - 1 ) = 2\sum_n a_n \sin^2 (nx /2) . $$
Then we divide the summation into two segments,
$$ g(x) = g_1 + g_2 = 2 (\sum_{n\leq M } + \sum_{n>M} ) a_n \sin^2 (nx/2). $$
We should choose $M$ so large that $|a_n /n^{-3/2} - \alpha |<\epsilon $ when $n>M $. The first part $g_1 $ is $O(x^2)$. The second part is bounded as
$$ (\alpha - \epsilon )\sum_{n>M} \frac{2 \sin^2 n x /2}{n^{3/2}} < g_2 < (\alpha + \epsilon )\sum_{n>M} \frac{2 \sin^2 n x /2}{n^{3/2}} . $$
Now I tend to approximate the summation as an integral,
$$ \sum_{n>M} \frac{2 \sin^2 n x /2}{n^{3/2}} \simeq x^{1/2}\int_{Mx}^\infty \frac{ 2 \sin^2 (y/2)}{y^{3/2}} dy \simeq x^{1/2}\int_{0}^\infty \frac{ 2 \sin^2 (y/2)}{y^{3/2}} dy . $$
As $\epsilon $ can be arbitrarily small, and for each $\epsilon$, we can find a $M $, I think the asymptotic formula is right. Moreover, the coefficient $c_1$ can be determined once the integral is done.
Can anyone give a more concise proof or point out the loopholes in my arguments?