# Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.sum_{n=2}^inftyfrac{1}{nsqrt{ln n}}

Question
Series

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
$$\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}$$

2021-01-17
Apply integral test and check weather series is convergeny or divergent.
If the term series $$a_n$$ can be represented by a positive, decreasing, continuous functionm $$f(n)$$ then,
If $$\lim_{R\to\infty}\int_a^R f(x)dx$$ exists then $$\sum_{n=1}^\infty a_n$$ converges.
If $$\lim_{R\to\infty}\int_a^R f(x)dx$$ does not exist, then $$\sum_{n=1}^\infty a_n$$ diverges.
Consider the given series:
$$S_n=\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}$$
Determine the convergence of the following Integral:
$$\int_2^\infty\frac{1}{x\sqrt{\ln x}}dx=\lim_{t\to\infty}\int_{2}^t\frac{1}{x\sqrt{\ln x}}dx$$
Apply u-substitution: $$u=\ln(x)$$
$$\int_2^t\frac{1}{x\sqrt{\ln x}}dx=\int_{\ln(2)}^{\ln(t)}\frac{1}{\sqrt{u}}du$$
$$=\int_{\ln(2)}^{\ln(t)}u^{-\frac{1}{2}}du$$
$$=[\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}]_{\ln(2)}^{\ln(t)}$$
$$=[2\sqrt{u}]_{\ln(2)}^{\ln(t)}$$
$$=2\sqrt{\ln(t)}-2\sqrt{\ln(2)}$$
Find whether series diverges.
Applying Limits:
$$\lim_{t\to\infty}\int_2^t\frac{1}{x\sqrt{\ln x}dx}=\lim_{t\to\infty}[2\sqrt{\ln(t)}-2\sqrt{\ln(2)}]$$
$$=[\infty-2\sqrt{\ln(2)}]$$
$$=\infty$$
=Diverges
Hence, $$\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}$$ diverges

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