limit of fraction with summation

hi,

it has been some time since i last saw calculus and I'm stuck at one excercise. The point is just to find limit of

$\displaystyle \frac{ln(x)}{\sum_{k=1}^n \frac{1}{k}} $

What I would do is just solving the summation as definite integral and then differentiating the fraction via l'Hospital rule. Is that an ok way to go?

Re: limit of fraction with summation

But the summation is in terms of k ,, and the numerator is in term of x !

Re: limit of fraction with summation

Quote:

Originally Posted by

**pyjong** The point is just to find limit of

$\displaystyle \frac{ln(x)}{\sum_{k=1}^n \frac{1}{k}} $

What I would do is just solving the summation as definite integral and then differentiating the fraction via l'Hospital rule. Is that an ok way to go?

This question does not make sense. Where is the limit? $\displaystyle x\to?$ or $\displaystyle n\to\infty~?$

**What is the exact wording of the question?**

Re: limit of fraction with summation

oh .. sorry my bad, there is supposed to be n instead of x and the limit is in +inf

Re: limit of fraction with summation

Quote:

Originally Posted by

**pyjong** oh .. sorry my bad, there is supposed to be n instead of x and the limit is in +inf

It is well known that the Harmonic series is divergent, so the denominator tends to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$, and the numerator also tends to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$. Since this goes to $\displaystyle \displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$, you should be able to apply L'Hospital's Rule.

Re: limit of fraction with summation

Quote:

Originally Posted by

**pyjong** oh .. sorry my bad, there is supposed to be n instead of x and the limit is in +inf

So it is $\displaystyle \lim _{n \to \infty }}\frac{{\ln (n)}}{{\sum\limits_{k = 1}^n {\frac{1}{k}} }}$

For the harmonic series let $\displaystyle {H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} $.

Re: limit of fraction with summation

There is another way to do it but it requires a piece of knowledge. The Euler-Mascheroni constant is defined as

$\displaystyle \gamma = \lim_{n\to \infty} (H_n - \ln n)$

where $\displaystyle H_n$ is the $\displaystyle n$-th Harmonic number. Now we have

$\displaystyle \lim_{n\to \infty} \frac{\ln n}{H_n} =\lim_{n\to \infty} \frac{\ln n - H_n + H_n}{H_n} = \lim_{n\to \infty} \frac{\frac{\ln n - H_n}{H_n} + 1}{1}$

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