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I am trying to calculate this fraction:

$\sum_{n=1}^{101} \frac{100 \choose n-1}{103 \choose n}$. I wonder if there is some identity about binomial coefficient that would be helpful here.

Yang
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    $$ \binom {a} {b}=\dfrac{a!}{b!(a-b)!}$$ – J. W. Tanner Jan 07 '20 at 23:38
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    I am not sure it helps but $\dfrac{100 \choose n-1}{103 \choose n} = \dfrac{100!,n!,(103-n)!}{(n-1)!,(101-n)!,103!} = \dfrac{n , (103-n),(102-n)}{101\times102\times 103}$ – Henry Jan 07 '20 at 23:39
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    Once simplified, this is just a sum over the values of some cubic polynomial in $n$. There are closed formulas for such things. – lulu Jan 07 '20 at 23:47

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As mentioned in an earlier comment we can simplify:

$$ \dfrac{\binom{100}{n-1}}{\binom{103}{n}} = \dfrac{n \, (103-n)\,(102-n)}{101\times102\times 103} $$

so our sum is:

$$ \dfrac{1}{{101\times102\times 103}}\sum_{n=1}^{101}(n^3-205n^2+10506n) $$

Next, we can utilize Faulhaber's formula to simplify to:

$$ \dfrac{1}{{101\times102\times 103}} \left( \left(\dfrac{n(n+1)}{2}\right)^2 - 205 \dfrac{n(2n+1)(n+1)}{6} + 10506\dfrac{n(n+1)}{2} \right) $$

Finally, substituting n = 101 gives us:

$$ \sum_{n=1}^{101} \frac{100 \choose n-1}{103 \choose n} = \dfrac{26}{3} $$

Chris Henson
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