Evaluate
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
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 $$\frac{5050 \ \color\red{\Gamma(1/50)} \ \Gamma(101)}{\color\red{50} \ \Gamma(5051/50)}\frac{\color\red{50} \ \Gamma(5101/50)}{\color\red{\Gamma(1/50)} \ \Gamma(102)}=\frac{5050\cdot5051}{50\cdot 101}=5051$$