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

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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