Calculate limit using Stolz-Cesàro theorem

Question

Can someone help me calculate this limit using the Stolz-Cesàro theorem? $\lim_{n\to \infty } \frac{1+\frac12+......+\frac1n}{\ln n}$

Answer 1

$\ln n$ is unbounded and increasing and hence we can use the theorem: $$ \displaystyle\lim_{n\to \infty } \frac{1+\frac12+......+\frac1n}{\ln n}=\displaystyle\lim_{n\to \infty } \frac{\frac1{n+1}}{\ln (n+1)-\ln n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac1{n+1}}{\ln \frac {(n+1)}n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac{-1}{(n+1)^2}}{\frac1{(n+1)}-\frac 1n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac{-1}{(n+1)^2}}{\frac{-1}{n(n+1)}}=1\\ $$

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This is intuitively clear because, the harmonic series can be written as: $$ 1+\frac12+......+\frac1n=\ln n+\gamma+\epsilon_n $$ where $\gamma$ is Euler constant and $\epsilon_n$ goes to zero as $n\to\infty$.