\sum _{n=1}^{\infty }\left(\frac{1}{8\sqrt{n}+9\left(\ln \left(n\right)\right)^2}\right)
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MathBot Answer:
The infinite series \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{8 \sqrt{n} + 9 \log{\left(n \right)}^{2}}\) diverges.
We can use Raabe-Duhamel's test to determine convergence of this series.
Raabe-Duhamel's test is an extension of the ratio test which can sometimes be used in cases where the ratio test is inconclusive.
Given a series \(\displaystyle\sum a_n\), define \(\rho = \displaystyle\lim_{n \to \infty} n\left(\frac{a_n}{a_{n+1}}-1\right)\)
•If \(\rho > 1\) the series converges.
•If \(\rho < 1\) the series diverges.
•If \(\rho = 1\) or the limit does not exist the test is inconclusive.
$$\begin{aligned}L &= \lim_{n \to \infty}\left|\frac{\frac{1}{8 \sqrt{n + 1} + 9 \log{\left(n + 1 \right)}^{2}}}{\frac{1}{8 \sqrt{n} + 9 \log{\left(n \right)}^{2}}}\right| \\ &= \lim_{n \to \infty} \frac{8 \sqrt{n} + 9 \log{\left(n \right)}^{2}}{8 \sqrt{n + 1} + 9 \log{\left(n + 1 \right)}^{2}} \\ &= \frac{1}{2}\end{aligned}$$Therefore the series diverges.