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Which among

$$ \left(2\,k+1 \atop j\right),~~j=1,3,5,...,2\,k+1 $$

has the larger order of infinity when $k\rightarrow\infty$? I am pretty sure that the largest order is reached around $j=k$ but I miss a formal proof.

AlmostSureUser
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    Note $\left(2,k+1 \atop k\right)=\left(2,k+1 \atop k+1\right)$ – Henry Sep 20 '14 at 09:50
  • "Large order of infinity" is not really a very precise way of describing the behaviour. While the largest binomial coefficients are indeed at $j=k$ and $j=k+1$, the ratio of $\binom{2k+1}{k-1}$ to $\binom{2k+1}{k}$ approaches $1$ as $k\to\infty$. – André Nicolas Sep 20 '14 at 10:02
  • To me this makes no sense. If you are letting $k$ vary, then how can you limit the choices for $j$ depending on $k$? It would seem that $j$ can be any odd number, in which case your expression is polynomial of degree $j$ in $k$; clearly the larger $j$, the faster this grows with $k$. – Marc van Leeuwen Sep 20 '14 at 10:13
  • Related: http://math.stackexchange.com/questions/722952/how-do-you-prove-n-choose-k-is-maximum-when-k-is-lceil-frac-n2-rceil – lab bhattacharjee Sep 20 '14 at 11:05

1 Answers1

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Hint: Simplify $$\dfrac{\left(2\,k+1 \atop j+1\right)}{\left(2\,k+1 \atop j\right)}$$ and find which values of $j$ make this greater than, equal to, or less than $1$.

Henry
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  • I cannot see how the question would be asking for the largest one of those binomial coefficients for fixed $k$, among the various choices for $j$. If that were the case, what is "order of infinity" and and $k\to\infty$ all about? (And why not just plainly ask what I said?) Hence I cannot see how this answers the question, or why it should be accepted by OP. – Marc van Leeuwen Sep 20 '14 at 10:16