Proof of \frac{1}{1-\ln 2} =4-\cfrac{8}{14-\cfrac{72}{30-\cfrac{288}{52-\cfrac{800}{\ddots}}}}

Suggested Proofs Proof of \frac{1}{1-\ln 2} =4-\cfrac{8}{14-\cfrac{72}{30-\cfrac{288}{52-\cfrac{800}{\ddots}}}}

1

6 months ago

Lezhe

open1

where for \(n\ge1\),
\[
a_n=3n^2+7n+4,\quad b_n=-2n^2(n+1)^2,\quad a_0=4.
\]

Proof:

1. Define convergents:
\[
\frac{P_n}{Q_n}=a_0+\cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cdots+\cfrac{b_n}{a_n}}}.
\]
Recurrence relations:
\[
P_{-1}=1,\;P_0=a_0=4,\quad Q_{-1}=0,\;Q_0=1,
\]
\[
P_n=a_nP_{n-1}+b_nP_{n-2},\quad Q_n=a_nQ_{n-1}+b_nQ_{n-2}.
\]

2. By induction one finds
\[
P_n=2^{\,n+1}(n+1)!(n+2)!.
\]

3. Define
\[
y_n=\frac{Q_n}{P_n},\quad \frac{P_n}{Q_n}=\frac{1}{y_n}.
\]
From recurrence we obtain
\[
z_n:=y_n-y_{n-1}=-\frac{b_nP_{n-2}}{P_n}(y_{n-1}-y_{n-2}).
\]
Explicit computation gives
\[
-\frac{b_nP_{n-2}}{P_n}=\frac{n}{2(n+2)}.
\]
So
\[
z_n=\frac{n}{2(n+2)}z_{n-1}.
\]

4. Initial conditions:
\[
y_0=\tfrac14,\quad y_1=\tfrac{7}{24},\quad z_1=\tfrac{1}{24}.
\]
Hence
\[
z_n=\frac{1}{2^{\,n+1}(n+1)(n+2)}.
\]

5. Thus
\[
y_n=\frac14+\sum_{k=1}^n \frac{1}{2^{\,k+1}(k+1)(k+2)}.
\]

6. Limit:
\[
y=\lim_{n\to\infty} y_n
=\frac14+\sum_{k=1}^\infty \frac{1}{2^{\,k+1}(k+1)(k+2)}.
\]

7. Evaluate the series:
\[
\frac{1}{(k+1)(k+2)}=\int_0^1 x^k(1-x)\,dx,
\]
so
\[
\sum_{k=1}^\infty \frac{1}{2^{\,k+1}(k+1)(k+2)}
=\tfrac12\int_0^1 \frac{x(1-x)}{2-x}\,dx.
\]
Simplifying the integral gives
\[
\int_0^1 \frac{x(1-x)}{2-x}\,dx=\tfrac32-2\ln 2.
\]
Therefore
\[
\sum_{k=1}^\infty \frac{1}{2^{\,k+1}(k+1)(k+2)}
=\tfrac34-\ln 2,
\]
and hence
\[
y=1-\ln 2.
\]

8. Conclusion:
\[
\lim_{n\to\infty}\frac{P_n}{Q_n}=\frac{1}{y}=\frac{1}{1-\ln 2}.
\]

Q.E.D.