Who's up to the challenge? 34

Let $a_n = \sqrt [ n ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }$ and $b_n = n$ . Then the limit can be written $\lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} \; .$

Since $b_n$ is a strictly monotone divergent sequence, we can apply Stolz–Cesàro theorem , that is, if the following limit exists, then it must be equal to the limit in question:

$\lim_{n\to\infty} \dfrac{ a_n}{b_n} =\lim_{n\to\infty} \dfrac{\sqrt [ n ]{ \Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }}n =\lim_{n\to\infty} \sqrt[n]{ \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right)}{n^n} }\; .$

And again, if we apply Stolz–Cesàro theorem again, that is, if the following limit exists, then it must be equal to the limit in question:

$\begin{aligned} &&\lim_{n\to\infty} \left [ { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right)\Gamma \left( \frac { 1 }{ n+1 } \right)}{(n+1)^{n+1}} } \div { \dfrac{\Gamma (1)\Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1 }{ 3 } \right) \cdots\Gamma \left( \frac { 1 }{ n } \right) }{n^n }} \right ] \\ &&= \lim_{n\to\infty} \left [ \Gamma\left( \dfrac1{n+1} \right) \cdot \dfrac{n^n}{(n+1)^{n+1} } \right ] \\ &&= \lim_{n\to\infty} \left [ (n+1) \cdot \Gamma\left( \dfrac1{n+1} \right) \cdot \left( \dfrac n{n+1} \right)^n \right ] \\ &&= \lim_{n\to\infty} \left [ \underbrace{(n+1) \cdot \Gamma\left( \dfrac1{n+1} \right)}_{a_n} \cdot \underbrace{\left( 1 - \dfrac1{n+1} \right)^{n+1}}_{b_n} \div \underbrace{\left( 1 - \dfrac1{n+1} \right)}_{c_n} \right ] \\ \end{aligned}$

By logarithmic differentation, it's easy to show that $b_n \to \dfrac1e$ and $c_n \to 1$ . We are left to evaluate $\displaystyle \lim_{n\to\infty} \left [ (n+1) \cdot \Gamma\left( \dfrac1{n+1} \right)\right]$ . Let $m = \dfrac1{n+1}$ , then as $n\to\infty$ , $m\to 0$ , the limit is equivalent to

$\lim_{m\to 0} m \Gamma(m)\; .$

If we represent the Gamma function in terms of its Weierstrass Representation , we have

$s \Gamma(s) = e^{-\gamma s } \prod_{k=1}^\infty e^{s/k} \left( 1 + \dfrac sk \right)^{-1}$

As $s\to 0$ , RHS tends to 1, this can be shown by logarithmic differentation. Thus $\displaystyle \lim_{m\to 0} m \Gamma(m) = 1$ , so $a_n \to 1$ .

Putting it all together, the limit in question is equal to $\displaystyle \lim_{n\to\infty} (a_n \times b_n \div c_n) = 1 \times \dfrac1e \div 1 = \dfrac 1e$ .

Hence, $A = 1, B = 1 \Rightarrow A+B = \boxed2$ .