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Since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, but $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, is there some $m \in \mathbb{R}$, with $1 < m < 2$ that defines the "boundary" between convergence and divergence for $\sum_{n=1}^{\infty} \frac{1}{n^m}$?

spacecowboy
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    A well-known result is that $p$-series is convergent iff $p>1$. This can be proved with integral test. And so for your question, there is no such $m$. – Angae MT Apr 24 '24 at 19:55
  • Where do you think the boundary is? EG What do you know about $ \sum 1/ n^{3/2} $? – Calvin Lin Apr 24 '24 at 19:55
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    More generally, there is no “smallest divergent series” (nor is there any “largest convergent series”). Indeed, for any integer $k\ge0$, the series $$\sum\frac1{n^{m_0}{(\log n)}^{m_1}{(\log\log n)}^{m_2}\cdots{(\underbrace{\log\log\cdots\log}_k,n)}^{m_k}}$$ is convergent if and only if $(m_0,m_1,m_2,\ldots,m_k)>(1,1,1,\ldots,1)$ for the lexicographic ordering. – nejimban Apr 24 '24 at 20:01
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    An older discussion here – Raymond Manzoni Apr 24 '24 at 20:19
  • Thank you, I always find that my intuition breaks down when working with infinities. – spacecowboy Apr 24 '24 at 21:07

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First note that the series you presented: $$ \sum_{n=1}^\infty \frac1{n^s}=\zeta(s) $$ Which converges $\forall s \ni \operatorname{ Re}(s)>1$