# does this sum have a limit?

Define the sequence given by the finite sum $$a_n:=\sum_{k=2}^{n+1}\binom{2k}k\binom{n+1}k\frac{k-1}{2^k\binom{4n}k}.$$

Questions.

(1) Is $0<a_n<1$?

(2) Does the limit $\lim_{n\rightarrow\infty}a_n$ exist? If yes, what is its value?

• Not having to deal with such sums in my work, it intrigues me how does one come about such a formula? – Asaf Karagila Jan 24 '17 at 7:47
• Is this a research-level question...? Or did this site devolve? – user541686 Jan 24 '17 at 9:44
• @AsafKaragila I don't know either, but I guess the reason is co.combinatorics. – Dirk Jan 24 '17 at 13:54
• @Dirk: Is co-combinatorics just a long way of saying "mbinatorics"? :) – Asaf Karagila Jan 24 '17 at 14:03
• @AsafKaragila Well, sometimes you encounter these kind of sums by counting the number of objects that you can construct somehow. We once met this one: $$\sum_{k=0}^n \binom{n,k} \binom{k}{\lfloor \frac{k}{2} \rfloor} 2^{n-k}.$$ (Although it turned out to be a rather simple binomial coefficient.) – Daniel Soltész Jan 26 '17 at 16:40

For each specific $k$ we have $$\binom{2k}k\binom{n+1}k\frac{k-1}{2^k\binom{4n}k}\to \binom{2k}k\frac{k-1}{8^k},$$ that suggests that the sum tends to $$\sum_{k=2}^\infty \binom{2k}k\frac{k-1}{8^k}=1-\frac{1}{\sqrt 2},$$ as $$\sum_{k= 0}^\infty \binom{2k}k x^k =(1-4x)^{-1/2};\,\\\sum_{k= 0}^\infty k\binom{2k}k x^k =(x\frac{d}{dx})(1-4x)^{-1/2}=2x(1-4x)^{-3/2};\\ \sum_{k= 0}^\infty (k-1)\binom{2k}k x^k =(6x-1)(1-4x)^{-3/2}=|_{x=1/8}-\frac1{\sqrt{2}}.$$

In order to justify this limit changes (limit of a sum is a sum of limits) we majorate each term in our sum. Say, $$\frac{\binom{n+1}k}{\binom{4n}k}=\frac{(n+1)n(n-1)\dots(n-k+2)}{4n(4n-1)(4n-2)\dots (4n-k+1)}<4^{-k}(1+\frac{1}n)(1+\frac{1}{4n-1})<3\cdot 4^{-k}.$$ this allows to estimate the total sum of summands with $k\geqslant k_0$ by $3\sum_{k\geqslant k_0} \binom{2k}k\frac{k-1}{8^k}$, that tends to 0 for large $k_0$. Therefore the limit of your sum is indeed a sum of limits, moreover, the estimates are very specific and should prove that $0<a_n<1$ also after checking some small values of $n$.

Not an answer. First, Mathematica says that $$a_n = \frac{-4 n \, _2F_1\left(\frac{1}{2},-n-1;-4 n;2\right)+n \, _2F_1\left(\frac{3}{2},-n;1-4 n;2\right)+\, _2F_1\left(\frac{3}{2},-n;1-4 n;2\right)}{4 n}.$$

Secondly, for all reasonable values of $n$ it seems that $a_n < -0.7,$ and, indeed, I conjecture that $$\lim_{n\rightarrow \infty} a_n = -\frac{1}{\sqrt{2}}.$$

Now, this is a response to the original form of this question, with the sum going from $0.$ The first two terms contribute $-1$ to the sum, so for the current version, the conjectured limit is $$1-\frac{1}{\sqrt{2}}.$$

• Do you mind altering your estimate to start at $k=2$? Sorry for the change made. If not, you can redefine and call it $b_n$ (in your solution) to start at $k=0$? – T. Amdeberhan Jan 24 '17 at 4:54
• @T.Amdeberhan Well, $k=1$ contributes $0,$ while $k=0$ contributes $-1...$ – Igor Rivin Jan 24 '17 at 4:55
• I know, this is just to avoid any confusion by potential readers. – T. Amdeberhan Jan 24 '17 at 4:56