Almost Perfect!

Using the identity

$\begin{aligned} \frac {k-1}k \sum_{j=0}^\infty \frac 1{j+x \choose k} & = \frac 1{x-1 \choose k-1} & \small \color{#3D99F6} \text{Putting }x=k=50 \\ \frac {50-1}{50} \sum_{j=0}^\infty \frac 1{j+50 \choose 50} & = \frac 1{50-1 \choose 50-1} \\ \implies \sum_{n=50}^\infty \frac 1{n \choose 50} & = \frac {50}{49} \end{aligned}$

$\implies a+b = 50+49 = \boxed{99}$

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Proof

The identity used above based on the more general identity below.

$\large \frac {k-1}k \sum_{j=0}^M \frac 1{j+x \choose k} = \frac 1{x-1 \choose k-1} - \frac 1{M+x \choose k-1}$

We can prove the identity above is true for $k \ge 2$ and all non-negative integers $M$ by induction.

For $M=\color{#D61F06}0$ ,

$\begin{aligned} \frac {k-1}k \sum_{j=0}^{\color{#D61F06}0} \frac 1{j+x \choose k} & = \frac {k-1}k \times \frac 1{x \choose k} \\ & = \frac {k-1}k \times \frac {k!(x-k)!}{x!} \\ & = \frac {(k-1)!(x-k)!}{x!}(k-1) \\ & = \frac {(k-1)!(x-k)!}{x!}(x - (x-k+1)) \\ & = \frac {(k-1)!(x-k)!}{(x-1)!} - \frac {(k-1)!(x-k+1)!}{x!} \\ & = \frac 1{x-1 \choose k-1} - \frac 1{{\color{#D61F06}0}+x \choose k-1}\end{aligned}$

Therefore, the identity is true for $M=\color{#D61F06}0$ .

Assuming the identity is true for $M$ , then:

$\begin{aligned} \frac {k-1}k \sum_{j=0}^{\color{#D61F06}M+1} \frac 1{j+x \choose k} & = \frac {k-1}k \sum_{j=0}^M \frac 1{j+x \choose k} + \frac {k-1}k \times \frac 1{{\color{#D61F06}M+1}+x \choose k} \\ & = \frac 1{x-1 \choose k-1} - \frac 1{M+x \choose k-1} + \frac {(k-1)!{(\color{#D61F06} M+1}+x-k)!}{({\color{#D61F06}M+1}+x)!}(k-1) \\ & = \frac 1{x-1 \choose k-1} - \frac 1{M+x \choose k-1} + \frac {(k-1)!{(\color{#D61F06} M+1}+x-k)!}{({\color{#D61F06}M+1}+x)!}({\color{#D61F06}M+1}+x - ({\color{#D61F06}M+1}+x - k+1)) \\ & = \frac 1{x-1 \choose k-1} - \frac 1{M+x \choose k-1} + \frac {(k-1)!{(\color{#D61F06} M+1}+x-k)!}{(M+x)!} - \frac {(k-1)!(M+2+x-k)!}{({\color{#D61F06}M+1}+x)!} \\ & = \frac 1{x-1 \choose k-1} - \frac 1{M+x \choose k-1} + \frac 1{M+x \choose k-1} - \frac 1{{\color{#D61F06}M+1}+x \choose k-1} \\ & = \frac 1{x-1 \choose k-1} - \frac 1{{\color{#D61F06}M+1}+x \choose k-1} \end{aligned}$

The identity is also true for $M+1$ and therefore it is true for all $M \ge 0$ .

For $M \to \infty$ , $\displaystyle \frac {k-1}k \sum_{j=0}^\infty \frac 1{j+x \choose k} = \frac 1{x-1 \choose k-1}$

heres a solution that utilizes beta function $\dfrac{1}{{n}\choose{50}} = \dfrac{50!(n-50)!}{n!} =\dfrac{\Gamma(51)\Gamma(n-49)}{\Gamma(n+1)}=\dfrac{50\Gamma(50)\Gamma(n-49)}{\Gamma(n+1)}=50\beta(50,n-49)$ we write the summation as $S=50\sum_{n=50}^\infty \beta(50,n-49)=50\sum_{n=50}^\infty \int_0^1 x^{n-50} (1-x)^{49} dx$ now we will swap the sum and integral, which is legal because the sum of $x^n$ is absolutely convergent over (0,1). $S=50\int_0^1 x^{-50} (1-x)^{49}\sum_{n=50}^\infty x^n dx=50\int_0^1 x^{-50} (1-x)^{49}x^{50}(1-x)^{-1} dx=50\int_0^1(1-x)^{48} dx$ using $\int_a^b f(x)dx=\int_a^b f(a+b-x) dx$ we have $S=50\int_0^1 x^{48} dx = \dfrac{50}{49}$ and the answer to the question is then trivial.