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o-na [289]
3 years ago
7

Use root test to determine the series is convergent/divergent or inconclusive

Mathematics
1 answer:
Alinara [238K]3 years ago
8 0

You can first condense the series to make it simpler to study. If <em>n</em> is odd, then <em>n</em> = 2<em>k</em> - 1, and if <em>n</em> is even, then <em>n</em> = 2<em>k</em> for some <em>k</em> ≥ 1. So for each <em>k</em>, you can pair up the <em>k</em>-th terms of the odd- and even-indexed series.

odd:

\dfrac1{3^{\frac{n+3}2}}=\dfrac1{3^{k+1}}

even:

\dfrac1{3^{\frac n2}}=\dfrac1{3^k}

So the series can be re-indexed as

\displaystyle\sum_{n=1}^\infty a_n=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty \left(\frac1{3^{k+1}}+\frac1{3^k}\right)=\sum_{k=1}^\infty\frac4{3^{k+1}}

By the root test, the series converges, since

\displaystyle\lim_{k\to\infty}\sqrt[k]{\left|\frac4{3^{k+1}}\right|}=\frac13\lim_{k\to\infty}\left(\frac43\right)^{\frac1k}=\frac13

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