Is it possible to write this in closed form: ∑k=0n(nk)log⁡((nk))

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Is it possible to write this in closed form:  ∑k=0n(nk)log⁡((nk))

Allison Compton · Accepted Answer

You may start by symmetrizing the summand to get∑k=0nk(nk)log⁡(nk)=n2∑k=0n(nk)log⁡(nk) (1)The terms in the sum on the right hand side of (1) are symmetric around n2 and concentrated near k≈n2, so replacing log⁡(nk) with log⁡(nn2) gives a reasonable approximation, and an upper bound. That is,n2∑k=0n(nk)log⁡(nk)≈n22nlog⁡(nn2)Using Stirling&#039;s formula gives another approximation (and upper bound)n2∑k=0n(nk)log⁡(nk)≈n22n[(n+12)log⁡(2)−log⁡(nπ)2]

enveradapb · Accepted Answer

If f(n) is your sum, then ef(n) becomes an integer product, say p(n), formed by multiplying each binomial coefficient (nk) to the power k⋅(nk). That is,  p(n)=ef(n)=∏k=0n(nk)k(nk)  The first few terms are  p(1)=1, p(2)=22, p(3)=39, p(4)=244312, p(5)=250575

RizerMix · Accepted Answer

Ok since ((n),(k))=((n),(n−k)) then we can large index terms with small index terms: so: ∑k=0nk((n),(k))log⁡((n),(k))=∑k=0n/2(k((n),(k))log⁡((n),(k))+(n−k)((n),(n−k))log⁡((n),(n−k))) =∑k=0n/2n((n),(k))log⁡((n),(k))=n∑k=0n/2((n),(k))log⁡((n),(k)) Now we just need to show the rest is bounded by 2n−1log⁡(2n−1)

Is it possible to write this in closed form: \sum_{k=0}^n((n),(k))\log(((n),(k)))

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