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


$\ln n$ 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 $n\to\infty$.

  • $\begingroup$ how did you came up with -1/(n+1)^2 ? $\endgroup$ – Filip Oct 15 '13 at 15:56
  • $\begingroup$ i mean how you cam up with this? : $\displaystyle\lim_{n\to \infty } \frac{***\frac{-1}{(n+1)^2}***}{\frac1{(n+1)}-\frac 1n}\\$ $\endgroup$ – Filip Oct 15 '13 at 16:01
  • 1
    $\begingroup$ 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 $n\to\infty$ $\endgroup$ – Arash Oct 15 '13 at 16:38
  • $\begingroup$ and how from ${\ln \frac {(n+1)}n}$ you came up with = $\frac {1}{(n+1)} - \frac 1n$ ?? $\endgroup$ – Filip Oct 15 '13 at 16:58
  • $\begingroup$ $\ln(\frac{x+1}{x})=\ln(x+1)-\ln x$ and I guess you know how to differentiate w.r.t. $x$. :-) $\endgroup$ – Arash Oct 15 '13 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.