# Calculate limit using Stolz-Cesàro theorem

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

## 1 Answer

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

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 $n\to\infty$.

• how did you came up with -1/(n+1)^2 ? – Filip Oct 15 '13 at 15:56
• i mean how you cam up with this? : $\displaystyle\lim_{n\to \infty } \frac{***\frac{-1}{(n+1)^2}***}{\frac1{(n+1)}-\frac 1n}\\$ – Filip Oct 15 '13 at 16:01
• This is result of applying L'Hopital rule. However another way is to say $\ln\frac{n+1}{n}=\ln(1+\frac 1n) \approx \frac 1n$ when $n\to\infty$ – Arash Oct 15 '13 at 16:38
• and how from ${\ln \frac {(n+1)}n}$ you came up with = $\frac {1}{(n+1)} - \frac 1n$ ?? – Filip Oct 15 '13 at 16:58
• $\ln(\frac{x+1}{x})=\ln(x+1)-\ln x$ and I guess you know how to differentiate w.r.t. $x$. :-) – Arash Oct 15 '13 at 17:01