Estimation of the middle combination number

Task number: 3853

Estimate the combination number \( \binom{2m}{m} \) using the factorial estimates proved in the lecture (see the hint).

  • Hint

    Use estimates: \( e\left(\frac{n}e\right)^n < n! < ne\left(\frac{n}e\right)^n \)

  • Solution

    We substitute into \(\binom{2m}m=\frac{(2m)!}{(m!)^2}\) and get:

    \(\frac{(2m)!}{(m!)^2}< \frac{2me\left(\frac{2m}e\right)^{2m}}{\left(e\left(\frac{m}e\right)^m\right)^2}= \frac{2m2^{2m}}{e} \),

    \( \frac{(2m)!}{(m!)^2}> \frac{e\left(\frac{2m}e\right)^{2m}}{\left(2me\left(\frac{m}e\right)^m\right)^2}= \frac{e2^{2m}}{4m^2} \).

    For \(m\ge 2\) are both estimetes worse (indeed by a linear factor) than the estimates straightforwardly deduced from the binomial theorem:

    \( \frac{2^{2m}}{2m+1}< \binom{2m}m<2^{2m}\)

Difficulty level: Easy task (using definitions and simple reasoning)
Routine calculation training
Cs translation
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