Determining convergence/divergence of a series All terms are real numbers.
$$c_0+c_1+\cdots+c_{100}=0,\ c_0\neq0$$
$$a_n=c_0\sqrt{n}+c_1\sqrt{n+1}+\cdots+c_{100}\sqrt{n+100}$$
$$b_n=\sqrt[4]{n^3}$$
The question is to determine convergence/divergence of
$$\sum_{n=1}^\infty \frac{a_n}{b_n}.$$
So far I have shown that $\sum_{n=1}^\infty a_n$ is convergent:
\begin{align*}
    \sum_{n=1}^\infty a_n&=(c_0+c_1\sqrt{2}+\cdots+c_{100}\sqrt{101})+(c_0\sqrt{2}+c_1\sqrt{3}+\cdots+c_{100}\sqrt{102})+\cdots\\
    &=c_0+\sqrt{2}(c_0+c_1)+\sqrt{3}(c_0+c_1+c_2)+\cdots+\sqrt{100}(c_0+c_1+\cdots+c_{99})+\sqrt{101}(c_0+c_1+\cdots+c_{100})+\cdots\\
    &=c_0+\sqrt{2}(c_0+c_1)+\sqrt{3}(c_0+c_1+c_2)+\cdots+\sqrt{100}(c_0+c_1+\cdots+c_{99})+(c_0+c_1+\cdots+c_{100})\sum_{n=101}^\infty\sqrt{n}\\
    &=c_0+\sqrt{2}(c_0+c_1)+\sqrt{3}(c_0+c_1+c_2)+\cdots+\sqrt{100}(c_0+c_1+\cdots+c_{99}).
\end{align*}
but I am lost after this since I cannot apply the direct comparison test.
Edit: The rearrangement above doesn't work by Riemann series theorem

Hints:

*

*Show that since $b_n = n^{3/4}$, if $|a_n| < C n^{-1/4 - \epsilon}$ for some constant $C$ and $\epsilon > 0$, then the sequence converges.


*Show that $ |a_n | < ( \frac{1}{2} \sum |c_i| ) ( \sqrt{n+100} - \sqrt{n}) < C n^{-1/2}$ for large enough $n$, and some constant $C$.


*Hence, the series converges.
