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I am trying to find a closed form for the following sum: $$ \sum_{k=0}^{n-1} \left( \frac{1}{(k+1)(n-k)} \cdot \binom{n+1}{k+1}^2 \right) $$

What I have tried so far

I tried to simplify the expression by shifting the index. Let $j = k+1$. The limits change from $0 \dots n-1$ to $1 \dots n$. The term $(n-k)$ becomes $(n-(j-1)) = (n-j+1)$.

The sum becomes: $$ S = \sum_{j=1}^{n} \frac{1}{j(n-j+1)} \binom{n+1}{j}^2 $$

I noticed that I can use partial fraction decomposition on the fraction: $$ \frac{1}{j(n-j+1)} = \frac{1}{n+1} \left( \frac{1}{j} + \frac{1}{n-j+1} \right) $$

Substituting this back into the sum: $$ S = \frac{1}{n+1} \sum_{j=1}^{n} \left( \frac{1}{j} + \frac{1}{n-j+1} \right) \binom{n+1}{j}^2 $$

Due to the symmetry of binomial coefficients where $\binom{n}{k} = \binom{n}{n-k}$, the two terms in the sum are actually identical. This simplifies the expression to:

$$ S = \frac{2}{n+1} \sum_{j=1}^{n} \frac{1}{j} \binom{n+1}{j}^2 $$

I am stuck at this point. Is there a known identity for $\sum \frac{1}{j} \binom{n}{j}^2$, or is there a better way to approach this?

Any help or hints would be appreciated.

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    $\begingroup$ What led you to consider this sum? The values of the sum are not integers, so I'm not sure if it's related to combinatorics. $\endgroup$ Commented Nov 21 at 12:42
  • $\begingroup$ WolframAlpha leads me to believe there's no nice closed form $\endgroup$ Commented Nov 21 at 12:47
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    $\begingroup$ There's no closed form $\endgroup$ Commented Nov 21 at 14:07
  • $\begingroup$ @SangchulLee (n+1)^2 times this sum appears after simplifying the following sum which appeared in a problem: $n \in \mathbb{N}, n \geq 2 \text{ şi } x = \sum_{k=0}^{n-1} \frac{\left(C_{n+1}^{k+1}\right)^4}{C_n^k C_n^{k+1}}$ $\endgroup$ Commented Nov 21 at 20:38

2 Answers 2

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Here I show how to evaluate $\sum_{j=1}^{n-1}\frac{1}{j+1} \binom{n}{j}^2$ (this is almost what you need $\sum_{j=1}^{n-1}\frac{1}{j} \binom{n}{j}^2$ and I guess these two sums are related in some way and one of them can be expressed in terms of another).

Lemma (as far as I know this is a classic fact and it is called Egorychev identity for binomials): $$\binom{n}{k}=\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{k+1}}dz$$

Proof: $$\frac{1}{2 \pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{k+1}}dz = res_{z=0} \frac{(1+z)^n}{z^{k+1}}=\frac{1}{k!} \frac{d^{k}}{dz^{k}}\Bigg| _{z=0} \Big( (1+z)^n\Big)=\frac{1}{k!} n(n-1)(n-2)...(n-k+1)(1+z)^{n-k} \Bigg| _{z=0}=\frac{n(n-1)(n-2)...(n-k+1)}{k!}=\binom{n}{k}$$

Now we apply Lemma in the following way:$$\sum_{j=1}^{n-1}\frac{1}{j+1} \binom{n}{j}^2=\sum_{j=1}^{n-1}\frac{1}{j+1} \frac{n! (n+1)}{j!(n-j)! (n+1)} \binom{n}{j}=\frac{1}{n+1} \sum_{j=1}^{n-1}\frac{(n+1)!}{(j+1)!(n-j)!} \binom{n}{j}=\frac{1}{n+1} \sum_{j=1}^{n-1}\binom{n+1}{j+1} \binom{n}{j}=\frac{1}{2\pi i(n+1)} \sum_{j=1}^{n-1} \int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{j+2}}dz \binom{n}{j}=\frac{1}{2\pi i(n+1)}\int_{|z|=\epsilon}\frac{(1+z)^{n+1}}{z^2} \sum_{j=1}^{n-1} \bigg(\frac{1}{z} \bigg)^j \binom{n}{j}dz=\frac{1}{2\pi i(n+1)}\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^2} \bigg( \Big(1+\frac{1}{z}\Big)^n -1-\frac{1}{z^n}\bigg)dz=\frac{1}{2\pi i(n+1)}\int_{|z|=\epsilon} \bigg[ \frac{\big(z+1 \big)^{2n+1}}{z^{n+2}} -\frac{(z+1)^{n+1}}{z^2}-\frac{(z+1)^{n+1}}{z^{n+2}}\bigg]dz=\frac{1}{n+1} \bigg[ \binom{2n+1}{n+1} -\binom{n+1}{1} -\binom{n+1}{n+1} \bigg]=\frac{1}{n+1} \bigg[\binom{2n+1}{n+1}-n-2 \bigg]$$

Thus $$\sum_{j=1}^{n-1}\frac{1}{j+1} \binom{n}{j}^2=\frac{1}{n+1} \bigg(\binom{2n+1}{n+1}-n-2 \bigg)$$

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    $\begingroup$ This doesn't really relate to OPs sum if i am not wrong. But the one you are solving can be done simply using Vandermode Convolution. $\endgroup$ Commented Nov 21 at 14:14
  • $\begingroup$ @匚ㄖㄥᗪ乇ᗪ how it can be done via Vandermonde Convolution? It looks like Vandermonde Convolution gives a closed form for $\sum_{j=0}^{n+1}\binom{n}{j} \binom{n+1}{j}=\sum_{j=0}^{n+1}\binom{n}{j} \binom{n+1}{j+1} \frac{j+1}{n+1-j}$, but not for $\sum_{j=0}^{n+1}\binom{n}{j} \binom{n +1}{j+1}$ $\endgroup$ Commented Nov 21 at 14:26
  • $\begingroup$ \begin{aligned} \sum_{j=1}^{n} \frac{\binom{n}{j}^2}{j+1} &= \sum_{j=1}^{n} \binom{n}{j} \frac{1}{n+1} \binom{n+1}{j+1} \\ &= \frac{1}{n+1} \sum_{j=1}^{n} \binom{n}{n-j} \binom{n+1}{j+1} \\ &= \frac{1}{n+1} \left[ \binom{2n+1}{n+1} - \binom{n}{n}\binom{n+1}{1} \right] \\ &= \frac{\binom{2n+1}{n+1}}{n+1} - 1 \end{aligned} $\endgroup$ Commented Nov 21 at 18:56
  • $\begingroup$ @匚ㄖㄥᗪ乇ᗪ how exactly did you get $\sum_{j=1}^{n} \binom{n}{n-j} \binom{n+1}{j+1} = \binom{2n+1}{n+1}-\binom{n}{n} \binom{n+1}{1}$? Vandermonde Convolution states that $\sum_{j=0}^{r} \binom{m}{j} \binom{l}{r-j}=\binom{m+l}{r}$; if you take $r=n+1$ and $m+l=2n+1$ to get $\binom{2n+1}{n+1}$ then it equals to $\sum_{j=0}^{n+1}\binom{m}{j} \binom{l}{n+1-j}$, then $m$ must be $n$ and hence $l$ must be $n+1$, we obtain $\sum_{j=0}^{n+1}\binom{n}{j} \binom{n+1}{n+1-j}=\sum_{j=0}^{n+1}\binom{n}{n-j} \binom{n+1}{j}$. This is $j$ instead of $j+1$ in the second binomial. What do you mean?.. $\endgroup$ Commented Nov 21 at 19:29
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$$S_n=\sum_{k=0}^{n-1} \frac{1}{(k+1)(n-k)} \, \binom{n+1}{k+1}^2= \frac{2}{n+1} \sum_{j=1}^{n} \frac{1}{j}\, \binom{n+1}{j}^2 $$ Using the gamma function $$S_n=2(n+1)\Big(\Gamma(n+1) \Big)^2\,\sum_{j=1}^{n}\frac 1 {j^3\,\Big(\Gamma(j)\,\Gamma(n+2-j)\Big)^2}$$

Using generalized hypergeometric functions $$S_n=2 (n+1) \, _4F_3(1,1,-n,-n;2,2,2;1)-\frac{2}{(n+1)^2}$$

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  • $\begingroup$ @DS. Thank you for pointing it $\endgroup$ Commented Nov 24 at 3:37

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