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Consider Pascal's triangle without the $1$s.

enter image description here

Let $S$ be the sum of squares of reciprocals.

$$S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{4^2}+\frac{1}{5^2}+\cdots$$

Does $S$ have a closed form?

Wolfram does not evaluate $S$.

Significance of $S$

$S$ is a single non-trivial number representing all of Pascal's triangle without the $1$s. (The sum of unsquared reciprocals diverges because the harmonic series diverges.)

Why I suspect $S$ has a closed form

I suspect $S$ has a closed form, because many other infinite sums of reciprocals in Pascal's triangle have closed forms. Examples:

  • $\sum\limits_{k=1}^\infty\binom{k+2}{k}^{-2}=\frac43\pi^2-13$ (source)
  • $\sum\limits_{k=1}^\infty\binom{k+3}{k}^{-2}=9\pi^2-\frac{355}{4}$ (source)
  • $\sum\limits_{k=0}^\infty\binom{2k}{k}^{-1}=\frac43+\frac{2\pi\sqrt3}{27}$ (source).
  • In Pascal's triangle without the outer two layers, the sum of (unsquared) reciprocals is $3/2$, as I show below.

Showing that $1<S<2$

Consider the color-coded Pascal's triangle below. (The green and orange numbers extend below the diagram infinitely).

enter image description here

Let $T$ be the sum of reciprocals of the blue and green numbers. I found that $f(n)=\sum\limits_{k=2}^\infty\binom{n+k}{k}^{-1}=\frac{2}{n^2-1}$ for $n>1$ (source). Then $T=\sum\limits_{n=2}^\infty f(n)=\sum\limits_{n=2}^\infty \frac{2}{n^2-1}=\frac32$ by telescoping.

So the sum of reciprocals of the green numbers is $\frac32-\left(\frac16+\frac{2}{10}+\frac{2}{15}+\frac{1}{20}+\frac{2}{21}+\frac{2}{35}\right)=\frac{67}{84}$.

So the sum of squares of reciprocals of the green numbers is between $0$ and $\left(\frac{67}{84}\right)^2\approx 0.6361961451$.

Using the fact that $\sum\limits_{k=2}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}-1$, the sum of squares of reciprocals of the orange and blue numbers is $2\left(\frac{\pi^2}{6}-1\right)-\left(\frac12\right)^2+\left(\frac{1}{6^2}+\frac{2}{10^2}+\frac{2}{15^2}+\frac{1}{20^2}+\frac{2}{21^2}+\frac{2}{35^2}\right)\approx 1.105202601$.

Therefore $1<S<2$.

Edit

In the comments, @Tyma Gaidash shows that $S=\frac23-\frac{2\pi}{9\sqrt3}+\int_0^1\frac{t(t^2-3t+1)\ln(t(1-t))}{(t^2-t+1)^3}dt\approx 1.1146$. I have asked about a closed form for the integral here.

Dan
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    $S=\sum_{k=1}^\infty\sum_{n>k}{n\choose k}^{-2}$. Perhaps it's possible to find a closed form for the inner sums first? Using partial fraction decomposition they can be written as a $\mathbb{Q}$-linear combination of sums of the form $\sum_{n>k}\frac1{(n-m)(n-j)}$, where $0\le m,j\le k-1$ are integers. – Mastrem Dec 16 '23 at 16:34
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    We use hypergeometric $_3 F_2$’s integral representation. Summing over the first sum gives $S=\sum\limits_{n=1}^\infty\frac{_3F_2(1,2,2;n+2,n+2;1)}{(n+1)^2}$ and applying the integral representation experimentally gives $S=\frac23-\frac{2\pi}{9\sqrt3}+\int_0^1\frac{t(t^2-3t+1)\ln(t(1-t))}{(t^2-t+1)^3}dt\approx 1.1146$ – Тyma Gaidash Dec 16 '23 at 18:10
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    Using the @ТymaGaidash's work alongside this result, your sum can be written in terms of the Clausen function as:

    $$S=\frac13+\frac{4}{3\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$

     – Zacky Dec 17 '23 at 04:16
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    Naturally one has to wonder whether the generalized variant also has a closed form: $$S(n)=\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{6^n}+\frac{1}{4^n}+\frac{1}{5^n}+\cdots$$ Perhaps if an integral is found then it is easily doable. – Zacky Dec 17 '23 at 04:35
  • I have asked about the general case. – Dan Dec 17 '23 at 22:59

1 Answers1

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From the comments:

@Mastrem notes that $S=\sum_{k=1}^\infty\sum_{n>k}{n\choose k}^{-2}$.

@Tyma Gaidash notes that, using hypergeometric ${}_3F_2$'s integral representation, summing over the first sum gives $S=\sum\limits_{n=1}^\infty\frac{_3F_2(1,2,2;n+2,n+2;1)}{(n+1)^2}$. Applying the integral representation experimentally gives $S=\frac23-\frac{2\pi}{9\sqrt3}+\int_0^1\frac{t(t^2-3t+1)\ln(t(1-t))}{(t^2-t+1)^3}\mathrm dt$ (which is supported by numerical evidence using A024746).

@Zacky uses a result to yield $S=\frac13+\frac{4}{3\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$ where $\operatorname{Cl}_2(x)$ is the Clausen function.

Edit:

Thanks to @Tyma Gaidash's comment to this answer, we have:

$$S=\frac13+\frac{4}{3\sqrt 3}G\approx 1.11463574623$$

where $G$ is Gieseking's Constant.

Dan
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