How can I calculate the following sum involving binomial terms: ∑k=0n(nk)(−1)k(k+1)2

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How can I calculate the following sum involving binomial terms:  ∑k=0n(nk)(−1)k(k+1)2

lirwerwammete9t · Accepted Answer

We have  1z∫0z∑k=0n(nk)skds=∑k=0n(nk)zkk+1  so that  −∫0z1t∫0t∑k=0n(nk)skdsdt=−∑k=0n(nk)zk+1(k+1)2  Setting z=−1 gives an expression for your sum,  ∑k=0n(nk)(−1)k(k+1)2=∫−101t∫0t∑k=0n(nk)skdsdt  Now, ∑k=0n(nk)sk=(1+s)n, so  ∑k=0n(nk)(−1)k(k+1)2=∫−101t∫0t(1+s)ndsdt  =1n+1∫−101t[(1+t)n+1−1]dt  =1n+1∫01un+1−1u−1du  =1n+1∫01∑k=0nukdu

Brenton Pennington · Accepted Answer

One application of the absorption identity gets one of the factors of k+1 out of the denominator:  ∑k=0n(nk)(−1)k(k+1)2=1n+1∑k=0n(n+1k+1)(−1)kk+1  =1n+1∑k=1n+1(n+1k)(−1)k+1k  By using the basic binomial coefficient recursion formula, we can make that happen.  Let f(n)=∑k=1n(nk)(−1)k+1k Then looking at the difference of f(n+1) and f(n) gives us  f(n+1)−f(n)=∑k=1n+1(n+1k)(−1)k+1k−∑k=1n(nk)(−1)k+1k  =∑k=0n(nk)(−1)kk+1  =1n+1∑k=0n(n+1k+1)(−1)k  =1n+1(1+∑k=0n+1(n+1k)(−1)k+1)  =1n+1  where in the last step we used the fact that the alternating sum of the binomial coefficients is 0.  Thus  f(n+1)=∑k=0n(f(k+1)−f(k))=∑k=0n1k+1=Hn+1  Therefore,

RizerMix · Accepted Answer

∑k=0n((n),(k))(−1)k(k+1)2 =−∑k=0n((n),(k))(−1)k∫01ln⁡(x)xkdx =−∫01ln⁡(x)∑k=0n((n),(k))(−x)kdx =−∫01ln⁡(x)(1−x)ndx =−ddμ∫01xμ(1−x)ndx|μ=0 =−ddμΓ(μ+1)Γ(n+1)Γ(μ+n+2)|μ=0=Hn+1n+1 Note that Γ(a+μ)=Γ(a)[1+(Ha−1−γ)μ+O(μ2)] such that =−Γ(μ+1)Γ(n+1)Γ(μ+n+2)=−n!(n+1)!×[1+[(1−γ)−(Hn+1−γ)]μ+O(μ2)] =−1n+1+Hnn+1μ+O(μ2) with H0=0

How can I calculate the following sum involving binomial terms: \sum_{k=0}^n((n),(k))\frac{(-1)^k}{(k+1)^2}

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