Answer: (a) \(\displaystyle \int_0^1 \ln(1+x)\,\mathrm dx \lt U_n\), where

\(\displaystyle U_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n.\)

(b) A lower bound is

\(\displaystyle L_n=\frac1n\ln\frac{(2n-1)!}{n!}-\frac{n-1}{n}\ln n,\)

equivalently

\(\displaystyle L_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n-\frac{\ln 2}{n}.\)

(c) \(\displaystyle U_n-L_n=\frac{\ln 2}{n}\to 0\) as \(n\to\infty\).

The function \(y=\ln(1+x)\) is increasing on \([0,1]\) because

\(\displaystyle \frac{d}{dx}\ln(1+x)=\frac1{1+x}\gt 0.\)

So rectangles using right endpoints give an upper bound, and rectangles using left endpoints give a lower bound.

(a) Divide \([0,1]\) into \(n\) intervals of width \(\frac1n\).

Using right endpoints \(\frac1n,\frac2n,\dots,\frac{n}{n}\), the rectangle sum is

\(\displaystyle \frac1n\ln\left(1+\frac1n\right)+\frac1n\ln\left(1+\frac2n\right)+\cdots+\frac1n\ln\left(1+\frac{n}{n}\right).\)

Since the function is increasing,

\(\displaystyle \int_0^1 \ln(1+x)\,\mathrm dx\lt \frac1n\sum_{r=1}^n\ln\left(1+\frac rn\right).\)

Now

\(\displaystyle \ln\left(1+\frac rn\right)=\ln\left(\frac{n+r}{n}\right)=\ln(n+r)-\ln n.\)

Hence

\(\displaystyle \frac1n\sum_{r=1}^n\ln\left(1+\frac rn\right)=\frac1n\left(\ln(n+1)+\ln(n+2)+\cdots+\ln(2n)-n\ln n\right).\)

Combining logs,

\(\displaystyle =\frac1n\ln\frac{(n+1)(n+2)\cdots(2n)}{n^n}.\)

Since \((n+1)(n+2)\cdots(2n)=\frac{(2n)!}{n!}\), this becomes

\(\displaystyle \frac1n\ln\frac{(2n)!}{n!}-\ln n.\)

Therefore

\(\displaystyle \int_0^1 \ln(1+x)\,\mathrm dx\lt U_n\)

with

\(\displaystyle U_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n.\)

(b) For the lower bound, use left endpoints \(0,\frac1n,\frac2n,\dots,\frac{n-1}{n}\).

The first term is \(\ln(1+0)=0\), so

\(\displaystyle \int_0^1 \ln(1+x)\,\mathrm dx\gt \frac1n\ln\left(1+\frac1n\right)+\frac1n\ln\left(1+\frac2n\right)+\cdots+\frac1n\ln\left(1+\frac{n-1}{n}\right).\)

Thus

\(\displaystyle L_n=\frac1n\sum_{r=1}^{n-1}\ln\left(1+\frac rn\right).\)

Using \(\ln(1+r/n)=\ln(n+r)-\ln n\),

\(\displaystyle L_n=\frac1n\left(\ln(n+1)+\ln(n+2)+\cdots+\ln(2n-1)-(n-1)\ln n\right).\)

So

\(\displaystyle L_n=\frac1n\ln\frac{(2n-1)!}{n!}-\frac{n-1}{n}\ln n.\)

An equivalent form is

\(\displaystyle L_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n-\frac{\ln 2}{n}.\)

(c) Using

\(\displaystyle U_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n\)

and

\(\displaystyle L_n=\frac1n\ln\frac{(2n)!}{n!}-\ln n-\frac{\ln 2}{n},\)

we get

\(\displaystyle U_n-L_n=\frac{\ln 2}{n}.\)

Since \(\displaystyle \frac{\ln 2}{n}\to 0\) as \(n\to\infty\),

\(\displaystyle \lim_{n\to\infty}(U_n-L_n)=0.\)