Let $u_n=\sum_{k=1}^n\frac{1}{k}-\ln(n)$ and $v_n = \sum_{k=1}^n \frac{1}{k} - \ln(n+1)$.
Approximate the common limit of the 2 sequences with a $10^{-2}$ precision.
Using the fact that the 2 sequences are adjacent, I tried to first find starting from which rank $n$ we have $|u_n-v_n|\leq 10^-{2}$.
$$ \begin{align*} |u_n-v_n|\leq 0.01 &\implies |\ln(n+1)-\ln(n)|\leq0.01\\ &\implies \ln(1+\frac{1}{n})\leq 0.01\\ &\implies 1+\frac{1}{n}\leq e^{0.01}\\ &\implies n \geq \left\lfloor \frac{1}{e^{0.01}-1} \right\rfloor \end{align*} $$
Hence I took $n=100$. But computing $\sum_{k=1}^{100}\frac{1}{k}-\ln(100)$ is rather tedious.
Is this how I'm supposed to do it or are there any simpler ways?