Finding the limit $\lim_{n \to \infty}{\frac{\Sigma_{0}^{n}(1/n)}{\ln(n)}}$ Let
$$
\lim_{n \to \infty}{\frac{
\sum_{1}^{n}(\frac{1}{n})}{\ln(n)}}
$$
Please provide some hint or a solution.
Thanks!
 A: by Stolz Cesaro theorem,    
it's $~\displaystyle\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=1}^{n+1}\dfrac{1}{k}-\sum_{k=1}^{n}\dfrac{1}{k}}{\ln(n+1)-\ln n}=\lim_{n\to\infty}\dfrac{1}{\dfrac{n+1}{n}\ln\left(1+\dfrac{1}{n}\right)^{n}}=1$
A: Using Stolz–Cesàro theorem you get that
$$\lim\limits_{n\to\infty} \frac{\sum_{k=1}^n\frac1k}{\ln n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln (n+1)-\ln n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln \frac{n+1}n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln (1+\frac1n)}.$$
(Of course, you should also verify that the assumptions of Stolz-Cesaro theorem are fulfilled.)
Now you can use the well-known limit $\lim\limits_{t\to0} \frac{\ln(1+t)}t=1$.

I will also mention that there are two equivalent formulations of Stolz-Cesaro theorem.
The same problem appeared in the book
Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series as Problem 2.3.21. The problem is stated on p.39 and solved on p.186. 
A: The limit is $1$; they differ by a constant: See Mascheroni constant
A: By the integral test we can consider the form for the numerator

$$ \lim_{n\to \infty}\frac{\int_{1}^{n}\frac{dx}{x}}{\ln(n)}= \lim_{n\to \infty}\frac{F(n)}{\ln(n)} $$

Now, we can apply L'hopital's rule to the above which gives
$$ \lim_{x\to \infty}\frac{1/n}{1/n}=1. $$
Notes:

$$ \sum_{k=1}^{n}\frac{1}{k}\sim \int_{1}^{n}\frac{1}{x}dx .$$
$$ \lim_{n\to \infty }F(n) = \infty. $$

A: You can also notice that the numerator is the harmonic number which grows as fast as the logarithm of n. When n goes to infinity, the limit of (H[n] - Log[n]) being the Euler–Mascheroni constant (as mentioned by Arjang). Then, your limit is 1.  
Have a look at http://en.wikipedia.org/wiki/Harmonic_number
