Pi and e are fine together

Working through the first few $x_n$ , we find that $\displaystyle x_n = (n-1)!\pi + (n-1)!\sum_{k=0}^{n-1} \frac {\pi^k}{k!}$ (see Note ). Let us prove it by induction that the claim is true for all $n \ge 1$ .

1. For $n = 1$ , $x_1 = 0!\pi + \dfrac {\pi^0}{0!} = \pi + 1$ as given. Therefore, the claim is true for $n=1$ .

2. Assume that the claim is true for $n$ , then we have:

$\begin{aligned} \quad \quad \quad x_{n+1} & = \pi^n + nx_n \\ & = \pi^n + n\left((n-1)!\pi + (n-1)!\sum_{k=0}^{n-1} \frac {\pi^k}{k!}\right) \\ & = n!\pi + n!\sum_{k=0}^{n-1} \frac {\pi^k}{k!} + \pi^n \\ & = n!\pi + n!\sum_{k=0}^{n} \frac {\pi^k}{k!} \\ \implies x_{\color{#3D99F6}{n+1}} & = (\color{#3D99F6}{n+1}-1)!\pi + n!\sum_{k=0}^{(\color{#3D99F6}{n+1}-1)} \frac {\pi^k}{k!} \end{aligned}$

$\quad \quad$ The claim is also true for $n+1$ and hence true for all $n\ge 1$ .

Now, we have:

$\begin{aligned} P & = \lim_{m \to \infty} \prod_{n=1}^m \left(1 - \frac {\pi^n}{x_{n+1}}\right) & \small \color{#3D99F6}{\text{As }x_{n+1}=\pi^n + nx_n} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac {nx_n}{x_{n+1}} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac {n\left((n-1)!\pi + (n-1)!\sum_{k=0}^{n=1} \frac {\pi^k}{k!}\right)}{n!\pi + n!\sum_{k=0}^{n} \frac {\pi^k}{k!}} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac {\pi + \sum_{k=0}^{n-1} \frac {\pi^k}{k!}}{\pi + \sum_{k=0}^{n} \frac {\pi^k}{k!}} \\ & = \lim_{m \to \infty} \frac {\prod\color{#D61F06}{_{n=1}^m} \left(\pi + \sum_{k=0}^{\color{#D61F06}{n-1}} \frac {\pi^k}{k!}\right)}{\prod_{n=1}^m \left(\pi + \sum_{k=0}^{n} \frac {\pi^k}{k!}\right)} \\ & = \lim_{m \to \infty} \frac {\left(\pi + \frac {\pi^0}{0!}\right) \prod\color{#D61F06}{_{n=1}^{m-1}} \left(\pi + \sum_{k=0}^{\color{#D61F06}{n}} \frac {\pi^k}{k!}\right)}{\prod_{n=1}^m \left(\pi + \sum_{k=0}^{n} \frac {\pi^k}{k!}\right)} \\ & = \lim_{m \to \infty} \frac {\pi + 1}{\pi + \sum_{k=0}^{\color{#D61F06}{m}} \frac {\pi^k}{k!}} \\ & = \frac {\pi + 1}{\pi + e^\pi} \end{aligned}$

$\implies (a + b + c)^{a+b+c} = (0+1+1)^{0+1+1} = \boxed{4}$

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Note:

\(\begin{array} {} a_1 = \pi + 1 \\ a_2 = \pi + \pi + 1 = 2\pi + 1 \\ a_3 = \pi^2 + 4\pi + 2 \\ a_4 = \pi^3 + 3\pi^2 + 12\pi + 6 \\ a_5 = \pi^4 + 4\pi^3 + 12\pi^2 + 48\pi + 24 \\ a_6 = \pi^5 + 5\pi^4 + 20\pi^3 + 60\pi^2 + 240\pi + 120 \\ ... \end{array} \)

$\begin{aligned} \implies \ a_n & = \pi^{n-1} + \frac {(n-1)!}{(n-2)!}\pi^{n-2} + \frac {(n-1)!}{(n-3)!}\pi^{n-3} + ... + \frac {(n-1)!}{2!} \pi^2 + \color{#3D99F6}{2(n-1)! \pi} + (n-1)! \\ & = \frac {(n-1)!}{(n-1)!} \pi^{n-1} + \frac {(n-1)!}{(n-2)!}\pi^{n-2} + \frac {(n-1)!}{(n-3)!}\pi^{n-3} + ... + \frac {(n-1)!}{2!} \pi^2 + \color{#3D99F6}{\frac {(n-1)!}{1!}\pi} + \frac{(n-1)!}{0!} + \color{#3D99F6}{(n-1)!\pi} \\ \implies a_n & = (n-1)!\pi + (n-1)!\sum_{k=0}^{n-1} \frac {\pi^k}{k!} \end{aligned}$

$\displaystyle \prod _{ n=1 }^{ k }{ \left( 1-\frac { { \pi }^{ n } }{ { x }_{ n+1 } } \right) } =\quad \prod _{ n=1 }^{ k }{ \left( \frac { n{ x }_{ n } }{ { x }_{ n+1 } } \right) } =\frac { n!\times { x }_{ 1 } }{ { x }_{ k+1 } }$

$\displaystyle { x }_{ k+1 }={ \pi }^{ k }+k{ x }_{ k }={ \pi }^{ k }+{ k\cdot \pi }^{ k-1 }+k\cdot (k-1){ \cdot x }_{ k-1 }=...=\sum _{ n=2 }^{ k }{ \frac { n!\cdot { \pi }^{ k } }{ k! } } +n!\cdot { x }_{ 2 }=\sum _{ n=2 }^{ k }{ \frac { n!\cdot { \pi }^{ k } }{ k! } } +n!\cdot \pi +n!\cdot { x }_{ 1 }=\sum _{ n=2 }^{ k }{ \frac { n!\cdot { \pi }^{ k } }{ k! } } +n!\cdot \pi +n!\cdot \pi +n!=\sum _{ n=0 }^{ k }{ \frac { n!\cdot { \pi }^{ k } }{ k! } } +n!\cdot \pi$

$\displaystyle \lim _{ k\rightarrow \infty }{ \dfrac { k!\cdot (1+\pi ) }{ { x }_{ k+1 } } } =\lim _{ k\rightarrow \infty }{ \dfrac { k!\cdot (1+\pi ) }{ k!\cdot (\sum _{ 0 }^{ k }{ \dfrac { { \pi }^{ k } }{ k! } } +\pi ) } } =\lim _{ k\rightarrow \infty }{ \dfrac { (1+\pi ) }{ \sum _{ 0 }^{ k }{ \dfrac { { \pi }^{ k } }{ k! } } +\pi } } =\dfrac { 1+\pi }{ \lim _{ k\rightarrow \infty }{ \sum _{ 0 }^{ k }{ \dfrac { { \pi }^{ k } }{ k! } } } +\pi } =\dfrac { 1+\pi }{ \pi +{ e }^{ \pi } }$

$\displaystyle a=0$ , $\displaystyle b=1$ and $\displaystyle c=1$

$\displaystyle (a+b+c)^{a+b+c}=2^{2}=\boxed{4}$

We have,

$x_{n+1} = \pi^n + nx^n$

$\implies \dfrac{x_{n+1}}{n!} = \dfrac{\pi^n}{n!} + \dfrac{x^n}{(n-1)!}$

$\implies \dfrac{x_{n+1}}{n!} - \dfrac{x^n}{(n-1)!} = \dfrac{\pi^n}{n!}$

$\implies \dfrac{x_{r+1}}{r!} - \dfrac{x^r}{(r-1)!} = \dfrac{\pi^r}{r!}$

$\implies \sum_{r=1}^{n-1} \left(\dfrac{x_{r+1}}{r!} - \dfrac{x^r}{(r-1)!}\right) = \sum_{r=1}^{n-1} \dfrac{\pi^r}{r!} \quad \quad (*)$

Clearly $(*)$ telescopes, evaluating it, we have,

$x_n = (n-1)!\pi + (n-1)!\sum_{k=0}^{n-1} \dfrac {\pi^k}{k!}$

Now,

$\text{P} = \prod _{ k=1 }^{ n }{ \left( 1-\dfrac { { \pi }^{ k } }{ { x }_{ k+1 } } \right) } = \prod _{ k=1 }^{ n }{ \left( \dfrac { k{ x }_{ k } }{ { x }_{ k+1 } } \right) } =\dfrac { n!\times { x }_{ 1 } }{ { x }_{ n+1 } }$

where the last equality follows from telescoping the product. Putting the value of $x_{n+1}$ and letting $n \to \infty$ , we get,

$\text{P} = \dfrac { 1+\pi }{ \pi +{ e }^{ \pi } }$

$\implies (a+b+c)^{a+b+c} = \boxed{4}$