调和级数求和

Get The Closed Form of (sum_{1le k le n}k^2H_{n+k})

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1、Use The Formula: (sum uDelta v = uv - sum EvDelta u)

We have known that: (sum u(x)Delta v(x) = u(x)v(x) - sum Ev(x)Delta u(x))

So apply the formula to our problem: (sum_{1le k le n}k^2H_{n+k})

Let (u(x) = H_{n+x}) and (Delta v(x) = x^2)

We can get that: (Delta u(x) = dfrac{1}{n+x+1})， (v(x) = dfrac{1}{3}x(x-1)(x-2) + dfrac{1}{2}x(x-1) = dfrac{1}{6}x(x-1)(2x-1))

so :

(sum x^2H_{n+x} = dfrac{1}{6}x(x-1)(2x-1)H_{n+x} - sum_{0 le k le n}dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1})

Observe the above formula, to solve this problem, we need get the closed form of (sum_{0 le k le n}dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1}), this is our next step.

2、Get The Closed Form of (sum_{0 le k le n}dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1})

First of all ,w need to simply the general term formula : (dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1}). Here is the steps how I handle it:

[egin{align*} &dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1}\ &= dfrac{1}{6}x[(x+1+n)-n](2x+1)*dfrac{1}{n+x+1} \ &= dfrac{1}{6}(x(2x+1) - dfrac{nx(2x+1)}{n+x+1}) \ &= dfrac{1}{6}(x(2x+1) - dfrac{nx(2x+1+1+2n -(1+2n))}{n+x+1}) \ &= dfrac{1}{6}(x(2x+1) - dfrac{nx(2(n+x+1) -(1+2n))}{n+x+1}) \ &= dfrac{1}{6}(x(2x+1) - 2nx + dfrac{n(1+2n)x}{n+x+1}) \ &= dfrac{1}{6}(x(2x+1) - 2nx + dfrac{n(1+2n)(x+n+1-(n+1))}{n+x+1}) \ &= dfrac{1}{6}(x(2x+1) - 2nx + n(2n+1) -dfrac{n(1+2n)(n+1)}{n+x+1}) \ &= dfrac{1}{6}x(2x+1) - dfrac{1}{3}nx +dfrac{1}{6}n(2n+1)-dfrac{1}{6}n(n+1)(2n+1)*dfrac{1}{n+x+1} end{align*} ]

Now, we can easily get the sum of (dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1}) from 1 to n:

[egin{align*} &sum_{1 le x le n}dfrac{1}{6}x(x+1)(2x+1)*dfrac{1}{n+x+1}\ &= sum_{1 le x le n}[dfrac{1}{6}x(2x+1) - dfrac{1}{3}nx +dfrac{1}{6}n(2n+1)-dfrac{1}{6}n(n+1)(2n+1)*dfrac{1}{n+x+1} ]\ &= dfrac{1}{6}sum_{1 le x le n}x(2x+1) - dfrac{1}{3}nsum_{1 le x le n}x +dfrac{1}{6}sum_{1 le x le n}n(2n+1)-dfrac{1}{6}n(n+1)(2n+1)*sum_{1 le x le n}dfrac{1}{n+x+1} \ &= dfrac{1}{3}sum_{1 le x le n}x^2 + dfrac{1}{6}sum_{1 le x le n}x -dfrac{1}{3}nsum_{1 le x le n}x + dfrac{1}{6}n^2(2n+1) - dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\ &=dfrac{1}{18}n(n+1)(2n+1) + dfrac{1}{12}n(n+1) -dfrac{1}{6}n^2(n+1)+ dfrac{1}{6}n^2(2n+1)- dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\ &=dfrac{1}{18}n(n+1)(2n+1) + dfrac{1}{12}n(n+1) + dfrac{1}{6}n^3 - dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\ &= dfrac{1}{36}n(n+1)(4n+5) + dfrac{1}{6}n^3 -dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\ end{align*} ]

3、The Most Existing Moment !!!

As we have got the two parts of the final result, now we just need to sum them, and merge the expression carefully.

[egin{align*} sum_{1le k le n}k^2H_{n+k} &= sum_{1 le k < n+1}k^2H_{n+k}\ &= sum_1^{n+1}k^2H_{n+k}delta k\ &= dfrac{1}{6}k(k-1)(2k-1)H_{n+k}|_1^{n+1} - sum_1^{n+1}(dfrac{1}{6}k(k+1)(2k+1)*dfrac{1}{n+k+1}) delta k\ &=dfrac{1}{6}k(k-1)(2k-1)H_{n+k}|_1^{n+1} - sum_{1 le k le n}dfrac{1}{6}k(k+1)(2k+1)*dfrac{1}{n+k+1}\ &= dfrac{1}{6}n(n+1)(2n+1)H_{2n+1} - dfrac{1}{36}n(n+1)(4n+5) - dfrac{1}{6}n^3 + dfrac{1}{6}n(n+1)(2n+1)*(H_{2n+1} -H_{n+1})\ &= dfrac{1}{6}n(n+1)(2n+1)(H_{2n+1} -H_{n+1}) - dfrac{1}{36}n(n+1)(4n+5) - dfrac{1}{6}n^3 \ &= dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} + dfrac{2}{2n+1} - H_n - dfrac{1}{n+1})- dfrac{1}{36}n(n+1)(4n+5) - dfrac{1}{6}n^3 \ &= dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) + dfrac{1}{6}n- dfrac{1}{36}n(n+1)(4n+5) - dfrac{1}{6}n^3 \ &= dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) - dfrac{1}{36}n(10n^2+9n-1) end{align*} ]

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Finally we got the result :

[egin{align*} sum_{1le k le n}k^2H_{n+k} &= dfrac{1}{6}n(n+1)(2n+1)(2H_{2n} - H_n) - dfrac{1}{36}n(10n^2+9n-1) end{align*} ]