Problem source

For $n=1$, $2$, $3$, $\ldots\,$, let
\[
I_n = \int_0^1 {t^{n-1} \over \l t+1 \r^n} \, \mathrm{d} t \, .
\]
By considering the greatest value taken by $\displaystyle {t \over t+1}$ for $0 \le t \le 1$ show that $I_{n+1} < {1 \over 2} I_{n}\,$. Show also that $\; \displaystyle I_{n+1}= - \frac 1{\; n\,  2^n}   + I_{n}\,$. Deduce that $\; \displaystyle I_n < \frac1 {\; n \, 2^{n-1}}\,$.
Prove that
\[
\ln 2 = \sum_{r=1}^n {1 \over \; r\, 2^r} + I_{n+1}
\]
and hence show that ${2 \over 3} < \ln 2 < {17 \over 24}\,$.

Solution source

\begin{align*}
&& \frac{t}{t+1} &= 1 - \frac{1}{t+1} \geq \frac12 \\
\Rightarrow && I_{n+1} &= \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\
&&&=  \int_0^1\frac{t}{t+1} \frac{t^{n-1}}{(t+1)^{n}} \d t \\
&&&< \int_0^1\frac12\frac{t^{n-1}}{(t+1)^{n}} \d t \\
&&&= \frac12 I_n \\
\\
&& I_{n+1} &=  \int_0^1 \frac{t^{n}}{(t+1)^{n+1}} \d t \\
&&&= \left [ t^n \frac{(1+t)^{-n}}{-n} \right]_0^1 +\frac1n \int_0^1 n t^{n-1}(1+t)^{-n} \d t \\
&&&= -\frac{1}{n2^n} + I_n \\
\Rightarrow && \frac12 I_n &> -\frac1{n2^n} + I_n \\
\Rightarrow && \frac{1}{n2^{n-1}} &> I_n
\end{align*}

\begin{align*}
&& \ln 2 &= \int_0^1 \frac{1}{1+t} \d t \\
&&&= I_1 \\
&&&= \frac1{2} + I_2 \\
&&&= \frac1{2} + \frac{1}{2 \cdot 2^2} + I_3 \\
&&&= \sum_{r=1}^n \frac{1}{r2^r} + I_{n+1} \\
\\ 
&& \ln 2 &= \frac12 + \frac18 + \frac1{24} + I_4 \\
\Rightarrow && \ln 2 &> \frac12 + \frac18 + \frac1{24} = \frac{12+3+1}{24} = \frac{16}{24} = \frac23 \\
\Rightarrow && \ln 2 &= \frac12 + \frac18 + I_3 \\
&&&< \frac12 + \frac18  +\frac{1}{3 \cdot 4} \\
&&&< \frac{12}{24} + \frac{3}{24} + \frac{2}{24} = \frac{17}{24}
\end{align*}