How should I prove convergence using Cauchy sequences $$ \sum_{k=1}^{n} \frac{\sin({k^{3}+1)}}{(4k+1)(4k+5)} $$
I tried starting with the definition $\forall\varepsilon>0,\exists N\in\mathbb{N}\ \forall n,p\geq N: |x_{n+p}-x_n|<\varepsilon$
$$ \left|\frac{\sin(2)}{(5)(9)} +\frac{\sin(9)}{(9)(13)}+ ...+\frac{\sin({n^{3}+1)}}{(4n+1)(4n+5)}+...+\frac{\sin({(n+p)^{3}+1)}}{(4(n+p)+1)(4(n+p)+5)}-\left(\frac{\sin(2)}{(5)(9)} +\frac{\sin(9)}{(9)(13)}+ ...+\frac{\sin({n^{3}+1)}}{(4n+1)(4n+5)}\right)\right|.$$
thus, after subtracting terms $$\left|\frac{\sin({(n+1)^{3}+1)}}{(4n+5)(4n+9)}+...+\frac{\sin({(n+p)^{3}+1)}}{(4(n+p)+1)(4(n+p)+5)}\right|$$
Since $\:-1\leq \sin(x) \leq 1\:$ and $\:n^2 \lt (4n+5)(4n+9)$ then $$\left|\frac{\sin({(n+1)^{3}+1)}}{(4n+5)(4n+9)}+...+\frac{\sin({(n+p)^{3}+1)}}{(4(n+p)+1)(4(n+p)+5)}\right| \lt\sum_{}^{\infty}{\frac{1}{n^2}}.$$
How do we find $n$ if the result of our greatest sum $\sum_{}^{\infty}{\frac{1}{n^2}}$, equals $\frac{\pi^2}{6}$
You have $$\begin{eqnarray} |x_{n+p}-x_n| & = &\left|\sum_{k=n+1}^{n+p} \frac{\sin (k^2+1)}{(4k+1)(4k+5)}\right|\\ &\leqslant & \sum_{k=n+1}^{n+p} \left|\frac{\sin (k^2+1)}{(4k+1)(4k+5)}\right|\\ & \leqslant& \sum_{k=n+1}^{n+p} \frac{1}{(4k+1)(4k+5)}\\ &=& \frac14 \sum_{k=n+1}^{n+p} \left(\frac{1}{(4k+1)}-\frac{1}{(4k+5)}\right)\\ &=& \frac14 \left( \frac{1}{4n+5} -\frac{1}{4(n+p)+5} \right)\\ &\leqslant & \frac{1}{4(4n+5)} \end{eqnarray} $$ and so choosing $N>\frac{1}{16\epsilon}$ will provide an $N$ for you to use in proving that $\{x_n\}_{1}^{\infty}$ is Cauchy.