Double Sum

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#1 h Apr 25, 2025, 3:22 AM

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Let $\xi = \lim_{m \to\infty} \sum_{n=4}^{m} \sum_{k=2}^{n-2} \frac{1}{\binom{n}{k}}.$ If the value of $\xi$ can be written as $\frac{m}{n}$ where $m$ and $n$ are co-prime positive integers then compute the value of $m^3 + n^3.$

This post has been edited 1 time. Last edited by P162008, Apr 25, 2025, 3:23 AM
Reason: Typo

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#2 h Apr 25, 2025, 11:42 AM • 2 Y

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We have $\xi=\sum_{k=2}^\infty \sum_{n=k+2}^\infty {1\over {{n\choose k}}}$ . Now $\sum_s^\infty {1\over {{n\choose s}}}={s\over {s-1}}$ for $s\ge 2$ .
For small values of $s$ I can prove this with the partial fraction decomposition for ${1\over {x(x-1)\cdots (x-s+1)}}$ and telescoping, but I don't have a general proof yet.
If the formula is correct then we get $\xi=\sum_2^\infty ({1\over {k-1}}-{1\over {k+1}})=3/2$ .

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#3 h Apr 25, 2025, 3:29 PM • 2 Y

Y by P162008, Sedro

Okay, so I just proved the identity in the first line. Let's give an example first.
We have $\sum_4^\infty {1\over {n\choose 4}}=\sum_4^\infty {{24}\over {n(n-1)(n-2)(n-3)}}=-4\sum_4^\infty ({1\over n}-{3\over {n-1}}+{3\over {n-2}}-{1\over {n-3}})=-4(\sum_4^\infty 1/n-3\sum_3^\infty 1/n+3\sum_2^\infty 1/n-\sum_1^\infty 1/n)=$
$-4(0-3(1/3)+3(1/2+1/3)-(1+1/2+1/3))=4/3$ . Now we generalise.

We have ${{s!}\over {x\cdots (x-s+1)}}=\sum_0^{s-1} {{c_k}\over {x-k}}$ where $c_k:=(-1)^{k+s+1} s{{s-1}\choose k}$ . Hence $\sum_s^\infty 1/{n\choose s}=\sum_s^\infty \sum_0^{s-1} {{c_k}\over {n-k}}=\sum_{k=0}^{s-1} c_k \{\sum_{n=s}^\infty {1\over {n-k}}\}=\sum_{k=0}^{s-1} c_k \{ \sum_{n=s-k}^\infty {1\over n}\}=\sum_{k=0}^{s-1} c_{s-1-k} \{\sum_{n=k+1}^\infty {1\over n}\}=$
$s\sum_{k=0}^{s-1} (-1)^k {{s-1}\choose k}\{\sum_{n=k+1}^\infty {1\over n}\}$ .
Now all terms with $n\ge s$ cancel by telescoping hence $s\sum_{k=0}^{s-1} (-1)^k {{s-1}\choose k}\{\sum_{n=k+1}^{s-1} {1\over n}\}=s\sum_{k=0}^{s-1} (-1)^k {{s-1}\choose k}(H_{s-1}-H_k)=s\sum_{k=0}^{s-1} (-1)^{k+1} {{s-1}\choose k} H_k$ . So that's better, we have a finite sum now.
We have to prove that $\phi_s:=\sum_0^s (-1)^{k+1} {s\choose k}H_k =1/s$ .
Let's try induction: $\phi_{s+1}=\sum_0^{s+1} (-1)^{k+1} {{s+1}\choose k} H_k=\sum_0^{s+1} (-1)^{k+1} \{ {s\choose k}+{s\choose {k-1}}\}H_k=\phi_s+\sum_0^s (-1)^k {s\choose k} (H_k+{1\over {k+1}})=$
$\sum_0^s {{(-1)^k}\over {k+1}} {s\choose k}={1\over {s+1}}\sum_0^ s (-1)^k {{s+1}\choose {k+1}}={{-1}\over {s+1}}\sum_1^{s+1} (-1)^k {{s+1}\choose k}={1\over {s+1}}$ . So no induction, but we're done.

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