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Evaluate

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There seemed to be some problem with stackexchange's math rendering but Ian corrected whatever error was there in the expression.Thanks

$$5050 \frac {\left( \sum _{r=0}^{100} \frac {{100\choose r}}{50r+1}\cdot (-1)^r\right) - 1}{\left( \sum _{r=0}^{101}\frac{{101\choose r}}{50r+1} \cdot (-1)^r\right) - 1}$$

The original definite integral that led to this was

$$5050\frac{\int_0^1(1-x^{50})^{100} dx}{\int_0^1(1-x^{50})^{101} dx} $$

I used binomail to arrive at the above expression.

I already know a technique by definite integration but I want one by sequence and series.

The answer is 5051

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    $\begingroup$ Pls don't downwote. I am using mobile so no preview available and I am trying to find error $\endgroup$ Aug 3, 2014 at 12:19
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    $\begingroup$ Please see if I got the intended expression right. $\endgroup$
    – Ian Mateus
    Aug 3, 2014 at 12:22
  • $\begingroup$ Hello now the question is perfect please remove the downvotes $\endgroup$ Aug 3, 2014 at 12:36

2 Answers 2

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Use of sequence and series is not suggested; and a possible way is outlined, which is very long and useless, this is due to the fact of bounded multiplication of r with 50 in denominator, if the r would have been free, it would be easy to use this method to caluculate:

Brute Force

$$\frac{100\choose r}{50r+1}=1-50.\frac{r.{100\choose r}}{50r+1}$$ Use: $$r{100\choose r}=100.{99\choose{r-1}}$$ Then $$\frac{{99\choose{r-1}}}{50r+1}=\left[1-50.\frac{r.{99\choose {r-1}}}{50r+1}\right]$$ Again use: $$r{99\choose {r-1}}=99.{98\choose{r-2}}\text{ after }\frac{r.{99\choose {r-1}}}{50r+1}=\frac{(r-1).{99\choose {r-1}}}{50r+1}+\frac{{99\choose {r-1}}}{50r+1}$$ and so on...


The way you are mentioning- definite integration of $\beta$-function (though you may not be realising it is the $\beta$-function)- is the best method in my opinion.


You should use these(preferred but not necessary) while summing binomial coefficients: $$r{n\choose r}=n{{n-1}\choose{r-1}};\frac1r{n\choose r}=\frac1{n+1}{{n+1}\choose{r+1}}$$ Secondly, use partial fractions or partial rationals.

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  • $\begingroup$ Thank you. But I was thinking more of using $n\choose r = n\choose {r-1} + {n-1}\choose {r-1} $ in numerator and getting something to cut with the other $\endgroup$ Aug 3, 2014 at 13:12
  • $\begingroup$ @MeghParikh See my edit for summation. $\endgroup$
    – RE60K
    Aug 3, 2014 at 13:13
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This expression can be simplified using Beta and Gamma functions.

The sum in the numerator is \begin{align} \sum^{100}_{r=0}\binom{100}{r}\frac{(-1)^r}{50r+1} &=\sum^{100}_{r=0}\binom{100}{r}(-1)^r\int^1_0x^{50r}dx\\ &=\int^1_0(1-x^{50})^{100}dx\\ &=\frac{1}{50}\int^1_0u^{-49/50}(1-u)^{100}du\\ &=\frac{1}{50}B(1/50,101)\\ &=\frac{\Gamma(1/50) \ \Gamma(101)}{50 \ \Gamma(5051/50)} \end{align} Similarly, the sum in the denominator is \begin{align} \int^1_0(1-x^{50})^{101}dx &=\frac{\Gamma(1/50) \ \Gamma(102)}{50 \ \Gamma(5101/50)} \end{align} Assuming you meant $$5050\frac{\sum^{100}_{r=0}\binom{100}{r}\frac{(-1)^r}{50r+1}}{\sum^{101}_{r=0}\binom{101}{r}\frac{(-1)^r}{50r+1}}$$ instead, the above expression evaluates to $$\frac{5050 \ \color\red{\Gamma(1/50)} \ \Gamma(101)}{\color\red{50} \ \Gamma(5051/50)}\frac{\color\red{50} \ \Gamma(5101/50)}{\color\red{\Gamma(1/50)} \ \Gamma(102)}=\frac{5050\cdot5051}{50\cdot 101}=5051$$

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  • $\begingroup$ I don't want an integration technique. I want a series and sequences technique. Plus I don't know beta and gamma functions. $\endgroup$ Aug 3, 2014 at 12:48
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    $\begingroup$ @MeghParikh Unfortunately, using Beta integrals to compute this expression is the most viable method for me. Also, there might be a slight error in the expression above, the $-1$s in both numerator and denominator should not be there for the answer to be $5051$ $\endgroup$ Aug 3, 2014 at 13:13
  • $\begingroup$ My method was to use IBP in $I_{100}$ and taking 1 as the easily integrable function. We get relation betn $I100\ and I101$ $\endgroup$ Aug 3, 2014 at 13:16
  • $\begingroup$ Also if you use binomial then you have to put both limits 0 and 1.. Due to zero the -1 is left $\endgroup$ Aug 3, 2014 at 13:18
  • $\begingroup$ @MeghParikh Note that $$\left[\sum^{100}_{r=0}\binom{100}{r}\frac{(-1)^r}{50r+1}\right]-1=\sum^{100}_{r=1}\binom{100}{r}\frac{(-1)^r}{50r+1}\neq\int^1_0(1-x^{50})^{100}dx$$. For the equality to hold, the sum must start from $r=0$ (equivalently, remove the $-1$ term. $\endgroup$ Aug 3, 2014 at 13:26

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