Prove: $\lim_{n\to\infty} \frac{\sum^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$ I'm not completely new to convergent series and sequences, but proving the above gave me some troubles. Would appreciate some help :)
Show that:
$$\lim_{n\to\infty} \frac{\sum\limits^n_{k=1} \frac{1}{k}}{ \ln(n) } = 1$$
 A: By approximating the sum with integrals (see here),
$$
\ln(n+1)=\int_1^{n+1}\frac1x\mathrm dx\le\sum_{k=1}^n\frac1k\le 1+\int_1^n\frac1x\mathrm dx=1+\ln n
$$
since $1/k$ is decreasing for $k\ge 1$. We have that
$$
\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln(1+\frac1n)}{\ln n}\to1
$$
as $n\to\infty$ and we obtain the result.
A: One can also use the Stolz–Cesàro theorem. Write 
$$\frac{\sum_{k=1}^n \frac 1k}{\ln n} = \frac{a_n}{b_n}.$$
As 
$$\begin{split}
\frac{a_{n+1} - a_n}{b_{n+1} - b_n} &= \frac{\frac{1}{n+1}}{\ln (n+1)- \ln n} \\
&= \frac{1}{(n+1) \ln\left(\frac{n+1}{n}\right)} \\
&= \left(\ln\left(\left(1 + \frac 1n\right)^n\left(1 + \frac 1n\right) \right)\right)^{-1} \\
&\to (\ln (e))^{-1} = 1,
\end{split}$$
then 
$$\lim_{n\to \infty} \frac{a_n}{b_n} = 1.$$
A: Starting with
$$\lim_{n\to\infty} \frac{\sum^n_{k=1} \frac{1}{k}}{ \log(n) }$$
Notice that the numerator represents the Harmonic numbers, so we rewrite this as
$$\lim_{n\to\infty} \frac{H_n}{ \log(n) }$$
Now, $H_n \sim \log(n) + \gamma$ for big $n$, where $\gamma \approx 0.5772156649$ is the Euler–Mascheroni constant. Plugging this in, we get
$$\lim_{n\to\infty} \frac{\log(n) + \gamma}{ \log(n) }$$
$$ 1+\lim_{n\to\infty}\frac{\gamma}{ \log(n) }$$
$$= 1$$
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