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Determine the convergence or divergence of the series.

1+(1)/(root(3)(4))+(1)/(root(3)(9))+(1)/(root(3)(16))+(1)/(root(3)(25))+dots

Determine the convergence or divergence of the series. $\newline$ $1+\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{9}}+\frac{1}{\sqrt[3]{16}}+\frac{1}{\sqrt[3]{25}}+\dots$

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Multiplication with rational exponents

Full solution

Q. Determine the convergence or divergence of the series.

\newline

1+\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{9}}+\frac{1}{\sqrt[3]{16}}+\frac{1}{\sqrt[3]{25}}+\dots

Identify series type: Step $1$ : Identify the series type. $\newline$ We're dealing with the series $1 + 1/\sqrt[3]{4} + 1/\sqrt[3]{9} + 1/\sqrt[3]{16} + 1/\sqrt[3]{25} + \ldots$ $\newline$ This looks like a p-series where each term is of the form $1/\sqrt[3]{n^2}$ .
Compare with convergent series: Step $2$ : Compare to a known convergent series. $\newline$ For p-series, $\sum \frac{1}{n^p}$ converges if p > 1. Here, $p = \frac{2}{3}$ (since the cube root of $n^2$ is $n^{\frac{2}{3}}$ ).

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