Find if this series converges or diverges.
The ratio test gives me limit 1 so I tried with Raabe Duhamel's test but I am stuck .
https://i.stack.imgur.com/ETKSt.png
$$\sum\limits_{m = 1}^\infty \frac{1}{1+\sqrt{2}+\sqrt[3]{3}+\dots+\sqrt[m]{m}}$$
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2$n^{1/n}< 2$; use the Comparison Test. – David Mitra Nov 10 '19 at 10:48
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$$ \frac{1}{2*n} < x_n $$ therefore the series is divergent? – Anon Nov 10 '19 at 11:08
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Good tool for numerical testing: https://www.desmos.com/calculator/w5cqiyokhp – clathratus Dec 07 '19 at 20:39
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1Does this answer your question? Convergence or divergence of $\sum_{n \in \mathbb{N}^{*}} \left(\sum_{k=1}^n k^{\frac{1}{k}}\right)^{-1}$ – Martin R Feb 06 '24 at 17:31
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Note that $\sqrt[m]{m} < 2\ \forall m \in \mathbb{N}$. Now you have $$\sum\limits_{k=1}^m \sqrt[k]{k} < \sum\limits_{k=1}^m 2 = 2m$$ Therefore $$\sum\limits_{m=1}^\infty \frac{1}{\sum\limits_{k=1}^m \sqrt[k]{k}} > \sum\limits_{m=1}^\infty \frac{1}{2m} = \infty$$ Thus the series diverges.
Note: Induction or applying the binomial theorem for $(1+1)^m$ gives you $m<2^m$ and therefore $\sqrt[m]{m} < 2$
GhostAmarth
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