Determine if the following series converges absolutely, converges conditionally or diverges: $$\sum\cos(n)\frac{(n+1)^n}{n^{n+1}}.$$
The series $\sum\cos n $ is bounded and the sequence $a_n=1/n$ is monotonic and $a_n\to 0$ and therefore we get that $$\sum\frac{\cos n}{n}$$ converges using Dirichlet's test. The sequence $b_n=\frac{(n+1)^n}{n^n}=(1+1/n)^n$ is increasing and $b_n\to e$ and due to Abel's test we get that the series $$\sum \frac{\cos(n)(n+1)^n}{n^{n+1}}$$ converges. Let's assume by contradiction that $$\sum\cos(n)\frac{(n+1)^n}{n^{n+1}}$$ is absolutely convergent. We have $$\left|\cos(n)\frac{(n+1)^n}{n^{n+1}}\right|\geq \left|\cos^2(n)\frac{(n+1)^n}{n^{n+1}}\right|$$ and from the comparison test we get that $$\sum\cos(n)\frac{(n+1)^n}{n^{n+1}}=\sum\frac{1}{2}\frac{(n+1)^n}{n^{n+1}}+\frac{1}{2}\cos(2n)\frac{(n+1)^n}{n^{n+1}}$$ converges. One can show that $$\sum\frac{1}{2}\cos(2n)\frac{(n+1)^n}{n^{n+1}}$$ converges in a similar manner as $$\sum\cos(n)\frac{(n+1)^n}{n^{n+1}}$$ was shown to converge. We thus have that $$\sum\frac{(n+1)^n}{n^{n+1}}=\sum2\cos^2(n)\frac{(n+1)^n}{n^{n+1}}-\cos(2n)\frac{(n+1)^n}{n^{n+1}}$$ converges which is a contradiction because $$0\leq\frac{n^n}{n^{n+1}}\leq\frac{(n+1)^n}{n^{n+1}}$$ and $$\frac{n^n}{n^{n+1}}=\frac{1}{n}$$ but $\sum\frac{1}{n}$ diverges and therefore we get using the comparison test that $\sum\frac{(n+1)^n}{n^{n+1}}$ diverges. We thus conclude that $$\sum\cos(n)\frac{(n+1)^n}{n^{n+1}}$$ is not absolutely convergent but we showed that it is convergent and therefore it is conditionally convergent.
Is this correct? If so, is there an easier way?