On convergence of series of the generalized mean $\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s.$

Solution 1:

Answer: Using the power mean (generalized mean) inequality the convergence for $s<0$ follows easily from the convergence for $s>1$. This probably does not give optimal bounds for $I_s$.

It's more natural to set $s=1/t$, so that the integrands are generalized means that, for fixed $(a_n)$, are increasing in $t \in \mathbb R$ by the power mean inequality (which says exactly that). We know that $I_{1/t}$ converges for $0<t<1$ (i.e. $s>1$, by Hardy's inequality, see here) and thus for all $-\infty \leq t < 1$ by the comparison test.

We also have that:

• the supremum $S_t=\sup(I_{1/t}/\sum a_n)$ over all sequences is finite for all $t<1$
• $S_t$ is (monotonically) increasing in $t$
• $S_{-\infty}=1$ (take a decreasing sequence)
• $S_{-1}=2$ (see this question)
• $S_0 \leq e$ (Carleman's inequality)
• $S_t \leq (1-t)^{-1/t}$ for $0<t<1$ (Hardy's inequality, see this question)

It's natural to conjecture that $S_t=(1-t)^{-1/t}$ for all $t<1$, which is $1$ at $-\infty$ and $e$ at $0$.