Lagrange's mean value theorem says this…

(a) Proof. $\ln x$ is continuous on $[1, 1 + h]$ and differentiable on $(1, 1 + h)$, with derivative $\dfrac1x$. By the theorem there is a $c$ with $1 < c < 1 + h$ and $\dfrac1c = \dfrac{\ln(1 + h) - \ln 1}{h}$, that is $\ln(1 + h) = \dfrac hc$. Since $1 < c < 1 + h$, we have $\dfrac{1}{1 + h} < \dfrac1c < 1$. Multiplying by $h > 0$: $\dfrac{h}{1 + h} < \ln(1 + h) < h$. The inequalities are strict because $c$ lies strictly inside the interval.

(b) With $h = 0.2$: $\dfrac{0.2}{1.2} = \frac{1}{6} \approx 0.1667$ and $0.2$. Rounding the lower bound down and the upper bound up, to three decimals: $0.166 < \ln 1.2 < 0.200$. (Indeed $\ln 1.2 \approx 0.1823$.) Rounding $0.1667$ to the nearest, $0.167$, would also still sit below $\ln 1.2$ here, but rounding down is the safe habit for a lower bound.

(c) $\dfrac1c = \dfrac{\ln e - \ln 1}{e - 1} = \dfrac{1}{e - 1}$, so $c = e - 1 \approx 1.72$, which does lie in $(1, e)$.

(d) Proof that the sequence decreases. $a_{n+1} - a_n = \dfrac{1}{n + 1} - [\ln(n + 1) - \ln n] = \dfrac{1}{n + 1} - \ln\left(1 + \dfrac1n\right)$. The left inequality of (a) with $h = \dfrac1n$ gives $\ln\left(1 + \dfrac1n\right) > \dfrac{1/n}{1 + 1/n} = \dfrac{1}{n + 1}$. So $a_{n+1} - a_n < 0$ for every $n$.

(d) Proof that every term is positive. The right inequality of (a) with $h = \dfrac1j$ gives $\dfrac1j > \ln\left(1 + \dfrac1j\right) = \ln(j + 1) - \ln j$. Adding for $j = 1, 2, \dots, n$, the right side telescopes: $1 + \dfrac12 + \dots + \dfrac1n > \ln(n + 1) > \ln n$. So $a_n > 0$. A decreasing sequence that stays positive must settle down to a limit; this limit, about $0.5772$, is known as Euler's constant.

(e) $1 + \dfrac12 + \dfrac13 + \dfrac14 + \dfrac15 = \frac{137}{60} \approx 2.2833$ and $\ln 5 \approx 1.6094$, so $a_5 \approx 0.6739$, that is 0.67.