Does the following series converge absolutely, converge conditionally or diverge: $$\sum_{n=1}^{\infty} \sin(n)\frac{(2^n+5^n)(n+1)^2}{n!}+\frac{\cos(1/n)}{\sqrt n}.$$
Let $$a_n=\sin(n)\frac{(2^n+5^n)(n+1)^2}{n!}$$ and let $$b_n=\frac{(2^n+5^n)(n+1)^2}{n!}.$$ We have $$b_n=\frac{2^nn^2}{n!}+\frac{2^n2n}{n!}+\frac{2^n}{n!}+\frac{5^nn^2}{n!}+\frac{5^n2n}{n!}+\frac{5^n}{n!}.$$ Observe that $$\lim_{n\to\infty} \left|\frac{\frac{2^{n+1}(n+1)^2}{(n+1)!}}{\frac{2^nn^2}{n!}}\right|=\lim_{n\to\infty} \frac{2n+2}{n^2}=0>1$$
$$\lim_{n\to\infty} \left|\frac{\frac{2^{n+1}2(n+1)}{(n+1)!}}{\frac{2^n2n}{n!}}\right|=\lim_{n\to\infty} \frac{2}{n}=0>1$$
$$\lim_{n\to\infty} \left|\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}\right|=\lim_{n\to\infty} \frac{2}{n+1}=0>1$$ and therefore the series $$\sum_{n=1}^{\infty} \frac{2^nn^2}{n!},\sum_{n=1}^{\infty} \frac{2^n2n}{n!},\sum_{n=1}^{\infty} \frac{2^n}{n!}$$ all converge due to the ratio test. In a similar manner, we can show that the series $$\sum_{n=1}^{\infty} \frac{5^nn^2}{n!},\sum_{n=1}^{\infty} \frac{5^n2n}{n!},\sum_{n=1}^{\infty} \frac{5^n}{n!}$$ all converge. We thus conclude that $\sum_{n=1}^{\infty} b_n$ converges. We have $$0\leq |a_n|=\left|\sin(n)\frac{(2^n+5^n)(n+1)^2}{n!}\right|\leq\left|\frac{(2^n+5^n)(n+1)^2}{n!}\right|=b_n$$ and therefore $\sum_{n=1}^{\infty} |a_n|$ converges which gives that $\sum_{n=1}^{\infty} a_n$ converges.
Let $c_n=\frac{\cos(1/n)}{\sqrt n}\geq 0$ and let $d_n=\frac{1}{\sqrt n} >0$. We have $$\lim_{n\to\infty} \frac{c_n}{d_n}=\lim_{n\to\infty}\cos(1/n)=\cos(0)=1>0$$ and because $$\sum_{n=1}^{\infty} d_n= \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$$ diverges, we obtain using the limit comparison test that $\sum_{n=1}^{\infty} c_n$ diverges. The series $\sum_{n=1}^{\infty} a_n$ converges and the series $\sum_{n=1}^{\infty} c_n$ diverges and therefore the series $$\sum_{n=1}^{\infty}=a_n+c_n=\sum_{n=1}^{\infty} \sin(n)\frac{(2^n+5^n)(n+1)^2}{n!}+\frac{\cos(1/n)}{\sqrt n}$$ diverges.
Is this correct?