Prove that 1/(a+1) < ln (1+a)/a < dfrac1a for a>0

Prove that $\dfrac{1}{a+1} \lt \ln\dfrac{1+a}{a} \lt \dfrac1a$ for $a>0$.

Step 1: Recognise an increment

\[ \ln\frac{1+a}{a} = \ln\bigl(a+1\bigr)-\ln a, \]

which is the increment of $\ln x$ over the interval $\bigl[a,\,a+1\bigr]$ — of length exactly $1$.

Step 2: Apply Lagrange's theorem

For $a>0$ the logarithm is continuous on $[a,a+1]$ and differentiable inside, with $\bigl(\ln x\bigr)' = \dfrac1x$. So for some $\xi\in\bigl(a,a+1\bigr)$

\[ \ln\bigl(a+1\bigr)-\ln a = \frac{1}{\xi}\cdot1 = \frac{1}{\xi}. \]

The unit length of the interval is what makes the answer simply $1/\xi$.

Step 3: Bracket $1/\xi$

The function $1/x$ is strictly decreasing for $x>0$, so $a\lt \xi\lt a+1$ gives

\[ \frac{1}{a+1} \lt \frac{1}{\xi} \lt \frac1a, \]

and substituting the identity of Step 2 completes the proof.

Step 4: Verification

\[a = 1:\quad 0.5 \lt \ln2 = 0.69315 \lt 1 \;\;\checkmark\]

\[a = 2:\quad 0.33333 \lt \ln\frac32 = 0.40546 \lt 0.5 \;\;\checkmark\]

\[a = 100:\quad 0.0099010 \lt \ln\frac{101}{100} = 0.0099503 \lt 0.01 \;\;\checkmark\]