Does this series converge

Question

I need to check by Cauchy's convergence test if this series converges: $$a_{n}=\frac{\sin(5)}{1*2} - \frac{\sin(5^2)}{2*3}+\ldots+(-1)^{n+1}\frac{\sin(5^n)}{n(n+1)}$$ I started to value $$|a_{n+p} - a_{n}| < \epsilon$$ But don't know how to continue from: $$\Biggl|(-1)^{n+2}*\frac{\sin(5^{n+1})}{(n+1)(n+2)}+\ldots+(-1)^{n+p+1}\frac{\sin(5^{n+p})}{(n+p)(n+p+1)}\Biggr|=\ldots$$

Answer 1

Let's the call the expression you have obtained as $S$. Applying triangle inequality to $S$ and using the fact that $|\sin(x)|\leq 1$ we get, \begin{align*} S \leq \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)}+ \cdots + \frac{1}{(n+p)(n+p+1)} \end{align*} By using the well known identity$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, the above expression simplifies to \begin{align*} \frac{1}{n+1} - \frac{1}{n+2}+\cdots+\frac{1}{n+p}-\frac{1}{n+p+1} = \frac{p}{(n+1)(n+p+1)}. \end{align*} Now choose an $n$ such that $\frac{1}{n+1}<\frac{\epsilon}{2} $. This implies \begin{align*} &&0 < \frac{1}{n+p+1}\leq\frac{1}{n+1}<\frac{\epsilon}{2}\\ \implies && S\leq \frac{p}{(n+p+1)(n+1)}<\epsilon \end{align*} Hence the above series is convergent