HypeMyday3m

2022-09-24

Upperbound for $\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}$

It's easy to prove that following upperbound is true:

$\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}\le A\mathrm{ln}A$, where $\sum _{i=1}^{N}{a}_{i}=A$ and ${a}_{i}\ge 1$

I'm wondering, is there stronger upperbound?

It's easy to prove that following upperbound is true:

$\sum _{i=1}^{N}{a}_{i}\mathrm{ln}{a}_{i}\le A\mathrm{ln}A$, where $\sum _{i=1}^{N}{a}_{i}=A$ and ${a}_{i}\ge 1$

I'm wondering, is there stronger upperbound?

Katelyn Ryan

Beginner2022-09-25Added 11 answers

You can also use the following estimate

$\sum _{j=1}^{N}{a}_{j}\mathrm{ln}{a}_{j}\le A\mathrm{ln}(A-N+1).$

$\sum _{j=1}^{N}{a}_{j}\mathrm{ln}{a}_{j}\le A\mathrm{ln}(A-N+1).$

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