infinity

n=1

for absolute convergence, I tried to compare with n^8/n^9 but I couldn't tell whether 1/n is smaller than the series. Is there another method I can test for absolute convergence? Also, on what basis can I conclude it does not converge absolutely (in other words, if a test for absolute convergence leads to a conclusion of divergence, is it entirely divergent or can it still conditionally converge?), and move on to test for conditional convergence?

\(\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n(4n^8 + 4)}{8n^9 + 2}\).

This series is absolutely convergent if \(\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| < 1\).

Don't try using the ratio test - all this time wasted...

So \(\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{(-1)^n[4(n + 1)^8 + 4]}{8(n + 1)^9 + 2}}{\frac{(-1)^n(4n^8 + 4)}{8n^9 + 2}}\right|\)

\(\displaystyle = \lim_{n \to \infty}\frac{\frac{4(n + 1)^8 + 4}{8(n + 1)^9 + 2}}{\frac{4n^8 + 4}{8n^9 + 2}}\)

\(\displaystyle = \lim_{n \to \infty}\frac{[4(n + 1)^8 + 4](8n^9 + 2)}{(4n^8 + 4)[8(n + 1)^9 + 2]}\)

\(\displaystyle = \lim_{n \to \infty}\frac{[(n + 1)^8 + 1](4n^9 + 1)}{(n^8 + 1)[4(n + 1)^9 + 1]}\).

\(\displaystyle = \lim_{n \to \infty}\)

\(\displaystyle \frac{(n^8 + 8n^7 + 28n^6 + 56n^5 + 70n^4 + 56n^3 + 28n^2 + 8n + 1 + 1)(4n^9 + 1)}{(n^8 + 1)[4(n^9 + 9n^8 + 36n^7 + 84n^6 + 126n^5 + 126n^4 + 84n^3 + 36n^2 + 9n + 1) + 1]}\)

\(\displaystyle =\lim_{n \to \infty}\)

\(\displaystyle \frac{\left(1 + \frac{8}{n} + \frac{28}{n^2} + \frac{56}{n^3} + \frac{70}{n^4} + \frac{56}{n^5} + \frac{28}{n^6} + \frac{8}{n^7} + \frac{2}{n^8}\right)\left(4 + \frac{1}{n^9}\right)}{\left(1 + \frac{1}{n^8}\right)\left[4\left(1 + \frac{9}{n} + \frac{36}{n^2} + \frac{84}{n^3} + \frac{126}{n^4} + \frac{126}{n^5} + \frac{84}{n^6} + \frac{36}{n^7} + \frac{9}{n^8} + \frac{1}{n^9}\right) + \frac{1}{n^9}\right]}\)

\(\displaystyle = \frac{1 \cdot 4}{1 \cdot 4}\)

\(\displaystyle = 1\).