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I am trying to evaluate the following sum :

$S=\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose n} \frac{1}{k+2}$

which is the same as

$\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose k} \frac{1}{k+2}$

My approach so far has been aimed at converting this sum to a known form (given as a standard result in Integrals and series by Prudnikov et.al.)

$\displaystyle\sum_{k=1}^n(-1)^{k+1}{n\choose k} {{n+k}\choose k} \frac{1}{k}=\sum_{k=1}^n\frac{1}{k}$

I thought of combining the like terms ${n\choose k} {{n+k}\choose k}$ in the two series but that again depends on whether $n$ is even or odd. Any suggestion to proceed further ?

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3 Answers 3

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There is a neat trick. The shifted Legendre polynomials fulfill $$ Q_n(x)=P_n(2x-1)=\sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}x^k \tag{1}$$ hence our sum is given by $$ S_n=\sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\frac{1}{k+2}=\int_{0}^{1}x\, Q_n(x)\,dx \tag{2}$$ and $$\boxed{\quad S_0=\frac{1}{2},\qquad S_1=\frac{1}{6},\qquad S_{n\geq 2}=0\quad}\tag{3} $$ follow from $x=\frac{Q_0(x)+Q_1(x)}{2}$ and the orthogonality relations for Legendre polynomials.

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  • $\begingroup$ (+1) It might be nice to mention the value of the square integral of the Legendre Polynomials (at least for $P_0$ and $P_1$). $\endgroup$
    – robjohn
    Aug 29, 2016 at 8:02
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$$ \begin{align} &\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{n}\frac1{k+2}\\[6pt] &=\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\left(\frac1{k+1}-\frac1{(k+1)(k+2)}\right)\tag{1}\\[6pt] &=\frac1{n+1}\sum_{k=0}^n(-1)^{n+k}\binom{n+1}{k+1}\binom{n+k}{k}\\ &-\frac1{(n+1)(n+2)}\sum_{k=0}^n(-1)^{n+k}\binom{n+2}{k+2}\binom{n+k}{k}\tag{2}\\[6pt] &=\frac{(-1)^n}{n+1}\sum_{k=0}^n\binom{n+1}{n-k}\binom{-n-1}{k}\\ &-\frac{(-1)^n}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{n-k}\binom{-n-1}{k} \tag{3}\\[6pt] &=\frac{(-1)^n}{n+1}\binom{0}{n}-\frac{(-1)^n}{(n+1)(n+2)}\binom{1}{n}\tag{4}\\[6pt] &=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{} \frac12&\text{if }n=0\\ \frac16&\text{if }n=1\\ 0&\text{if }n\ge2 \end{array}\right.}\tag{5} \end{align} $$ Explanation:
$(1)$: $\frac1{k+2}=\frac1{k+1}-\frac1{(k+1)(k+2)}$ and $\binom{n+k}{n}=\binom{n+k}{k}$
$(2)$: $\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$ applied once to the first sum and twice to the second
$(3)$: $\binom{n+j}{k+j}=\binom{n+j}{n-k}$ and $\binom{n+k}{k}=(-1)^k\binom{-n-1}{k}$
$(4)$: Vandermonde's Identity
$(5)$: evaluate at $n=0$ and $n=1$, and for $n\ge2$, $\binom{0}{n}=\binom{1}{n}=0$

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    $\begingroup$ i admire your ability to perform many complicated summations by only using elementary calculus (+1) $\endgroup$
    – tired
    Aug 29, 2016 at 7:43
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Suppose we seek to evaluate

$$(-1)^n \sum_{k=0}^n (-1)^{k} {n\choose k} {n+k\choose k} \frac{1}{k+2}.$$

This is

$$\frac{(-1)^n}{n+1} \sum_{k=0}^n (-1)^{k} {n+1\choose k+1} {n+k\choose k} \frac{k+1}{k+2} \\ = \frac{(-1)^n}{n+1} \sum_{k=0}^n (-1)^{k} {n+1\choose k+1} {n+k\choose k} \left(1- \frac{1}{k+2}\right).$$

The first piece here is

$$\frac{(-1)^n}{n+1} \sum_{k=0}^n (-1)^{k} {n+1\choose k+1} {n+k\choose k}$$

and the second

$$ \frac{(-1)^n}{n+1} \sum_{k=0}^n (-1)^{k} {n+1\choose k+1} {n+k\choose k} \frac{1}{k+2} \\ = \frac{(-1)^n}{(n+1)(n+2)} \sum_{k=0}^n (-1)^{k} {n+2\choose k+2} {n+k\choose k}.$$

Re-write the first piece as

$$\frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n+1} (-1)^{k} {n+1\choose k} {n+k-1\choose k-1}$$

Introduce

$${n+k-1\choose k-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k}} (1+z)^{n+k-1} \; dz.$$

This vanishes when $k=0$ so we may lower the index to zero to get for the sum

$$\frac{(-1)^{n+1}}{n+1}\frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{n-1} \sum_{k=0}^{n+1} {n+1\choose k} (-1)^k \frac{(1+z)^k}{z^k} \; dz \\ = \frac{(-1)^{n+1}}{n+1}\frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{n-1} \left(1-\frac{1+z}{z}\right)^{n+1} \; dz \\ = \frac{(-1)^{n+1}}{n+1}\frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{n-1} \frac{(-1)^{n+1}}{z^{n+1}} \; dz.$$

This is $$\frac{(-1)^{n+1}}{n+1}(-1)^{n+1} [z^n] (1+z)^{n-1} = 0$$

for $n\ge 1$. For $n=0$ we get

$$\frac{(-1)^{n+1}}{n+1}(-1)^{n+1} [z^0] \frac{1}{1+z} = 1.$$

Re-write the second piece as

$$\frac{(-1)^n}{(n+1)(n+2)} \sum_{k=2}^{n+2} (-1)^{k} {n+2\choose k} {n+k-2\choose k-2}.$$

Introduce

$${n+k-2\choose k-2} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k-1}} (1+z)^{n+k-2} \; dz.$$

This vanishes when $k=1$ and $k=0$ so we may lower the limit of the sum to zero, getting for the sum

$$\frac{(-1)^n}{(n+1)(n+2)} \frac{1}{2\pi i} \int_{|z|=\epsilon} z (1+z)^{n-2} \sum_{k=0}^{n+2} {n+2\choose k} (-1)^k \frac{(1+z)^k}{z^k} \; dz \\ = \frac{(-1)^n}{(n+1)(n+2)} \frac{1}{2\pi i} \int_{|z|=\epsilon} z (1+z)^{n-2} \left(1-\frac{1+z}{z}\right)^{n+2} \; dz \\ = \frac{(-1)^n}{(n+1)(n+2)} \frac{1}{2\pi i} \int_{|z|=\epsilon} z (1+z)^{n-2} \frac{(-1)^{n+2}}{z^{n+2}} \; dz \\ = \frac{(-1)^n}{(n+1)(n+2)} \frac{1}{2\pi i} \int_{|z|=\epsilon} (1+z)^{n-2} \frac{(-1)^{n+2}}{z^{n+1}} \; dz.$$

This is $$\frac{(-1)^n}{(n+1)(n+2)} (-1)^{n+2} [z^n] (1+z)^{n-2} = 0$$

for $n\ge 2.$ For $n=0$ we get

$$\frac{(-1)^n}{(n+1)(n+2)} (-1)^{n+2} [z^0] \frac{1}{(1+z)^2} = \frac{1}{2}.$$

For $n=1$ we get

$$\frac{(-1)^n}{(n+1)(n+2)} (-1)^{n+2} [z^1] \frac{1}{1+z} = - \frac{1}{6}.$$

Collecting everything we get $1/2$ for $n=0$ and $1/6$ for $n=1$ and zero otherwise.

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  • $\begingroup$ (+1) A tour de force. Nice job as always, but I still prefer shifted Legendre polynomials :D $\endgroup$ Aug 26, 2016 at 21:45
  • $\begingroup$ (+1) To you as well. The above what I wrote is only moderately pretty as we have repetition (two very similar computations). I guess it can serve as a basic example of the Egorychev method, possibly inducing the reader to want to learn more about it. $\endgroup$ Aug 26, 2016 at 22:27
  • $\begingroup$ (+1) The idea is similar to mine, but the integrals replace negative binomial coefficients and Vandermonde's Identity. $\endgroup$
    – robjohn
    Aug 29, 2016 at 7:50

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