test the series for convergence or divergence. \n∑(n = 1 to ∞) (7 - cos(n)) / √n\nconvergent\ndivergent\nneed…

Explanation:

Step1: Use the comparison - test

We know that $- 1\leqslant\cos(n)\leqslant1$. Then $7-\cos(n)\geqslant7 - 1=6$. So, $\sum_{n = 1}^{\infty}\frac{7-\cos(n)}{\sqrt{n}}\geqslant\sum_{n = 1}^{\infty}\frac{6}{\sqrt{n}}=6\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$.

Step2: Recall the p - series test

The p - series $\sum_{n = 1}^{\infty}\frac{1}{n^{p}}$ converges if $p>1$ and diverges if $p\leqslant1$. For the series $\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}=\sum_{n = 1}^{\infty}\frac{1}{n^{\frac{1}{2}}}$, where $p=\frac{1}{2}\leqslant1$, so $\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ diverges.

Step3: Conclusion

Since $\sum_{n = 1}^{\infty}\frac{7-\cos(n)}{\sqrt{n}}\geqslant6\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ and $6\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}$ diverges, by the comparison - test, the series $\sum_{n = 1}^{\infty}\frac{7-\cos(n)}{\sqrt{n}}$ diverges.

Answer:

divergent