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Is the ratio test for convergence applicable to the below series:

$$\sum_{n=1}^\infty \frac{n^3+1}{\sqrt[3]{n^{10} + n}}$$

I already know that the series diverge. I want to confirm if the ratio test is applicable or not?

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    what is your answer? – 2017-02-28
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    I dont know the answer. The question comes from a MCA question with 10 options, and I have to choose the ones that fail ratio test or root test for convergence – 2017-02-28
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    The ratio test is inconclusive if the ratio goes to 1. – 2017-02-28

4 Answers 4

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Let's compute the ratio

$${a_{n+1}\over a_n}={(n+1)^3+1\over n^3+1}\cdot {\sqrt[3]{(n+1)^{10}+n+1}\over \sqrt[3]{n^{10}+n}}\sim{n^{1\over 3}\over(n+1)^{1\over3}}\to 1$$

We cannot conclude with the ratio test

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hint: Compare your series with $\displaystyle \sum_{n = 1}^{\infty} \dfrac{1}{n^{\frac{1}{3}}}$, which diverges.

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    I understand that the series diverge. The question is ratio test applicable here? – 2017-02-28
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    The ratio test is not useful here since its $1$. – 2017-02-28
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What about of the integral test?

We have $$\frac{n^3+1}{\sqrt[3]{n^{10}+n}}> \frac{n^3+1}{\sqrt[3]{n^{10}+n^{10}}}= \frac{n^3+1}{2n^{10/3}}$$ Now, the integral $$\int_1^{\infty}\frac{x^3+1}{2x^{10/3}}dx$$ diverges. Then, the given series diverges too.

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If the limit of the ratio $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$$ Then the Ratio Test is Inconclusive. The test does not tell you anything about the series. The series may diverge or converge conditionally or absolutely.

As such, it would not be correct to say that the series fails the ratio test. It fails when the above limit is strictly greater than $1$.