# limit of fraction with summation

• Aug 25th 2012, 06:37 AM
pyjong
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?
• Aug 25th 2012, 06:42 AM
zaidalyafey
Re: limit of fraction with summation
But the summation is in terms of k ,, and the numerator is in term of x !
• Aug 25th 2012, 06:48 AM
Plato
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?
• Aug 25th 2012, 06:51 AM
pyjong
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
• Aug 25th 2012, 07:08 AM
Prove It
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.
• Aug 25th 2012, 07:13 AM
Plato
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 $\displaystyle {H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}}$.
• Aug 25th 2012, 09:15 AM
Vlasev
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

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

What's the next step? If you want to see, click "show"
Spoiler:
Now, as $\displaystyle n\to \infty$ the numerator in the inner fraction tends to $\displaystyle -\gamma$ and the denominator tends to $\displaystyle +\infty$. Therefore, the limit is

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