Finding the limit $\lim_{n \to \infty}{\frac{\Sigma_{0}^{n}(1/n)}{\ln(n)}}$

Question

Let

$$ \lim_{n \to \infty}{\frac{ \sum_{1}^{n}(\frac{1}{n})}{\ln(n)}} $$

Please provide some hint or a solution. Thanks!

You may mean $\sum_{k=1}^n \frac{1}{k}$ on top. If so, use the fact that the sum is close to $\int_1^n \frac{dx}{x}$. For precision you need a $2$-sided approximation, so you can use Squeezing. — André Nicolas, Nov 21 '13 at 07:43
Some users mentioned in comments/answers approximations of the sum by an integral (Euler's-Mascheroni constant). You can find several proofs on this site, for example: http://math.stackexchange.com/questions/451558/how-to-find-the-sum-of-this-series-1-frac12-frac13-frac14-do/ http://math.stackexchange.com/questions/344314/showing-that-lim-n-to-infty-sumn-k-1-frac1k-lnn-0-5772-ldots and http://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n — Martin Sleziak, Nov 21 '13 at 08:08

score 4 · Answer 1 · answered Nov 21 '13 at 07:48

4

by Stolz Cesaro theorem,

it's $~\displaystyle\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=1}^{n+1}\dfrac{1}{k}-\sum_{k=1}^{n}\dfrac{1}{k}}{\ln(n+1)-\ln n}=\lim_{n\to\infty}\dfrac{1}{\dfrac{n+1}{n}\ln\left(1+\dfrac{1}{n}\right)^{n}}=1$

answered Nov 21 '13 at 07:48

chloe_shi

2,855

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

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.

score 0 · Answer 3 · edited Nov 21 '13 at 07:44

0

The limit is $1$; they differ by a constant: See Mascheroni constant

edited Nov 21 '13 at 07:44

Mikasa

67,374

answered Nov 21 '13 at 07:39

jimjim

9,675

score 0 · Answer 4 · edited Jun 12 '20 at 10:38

0

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:

$$ \sum_{k=1}^{n}\frac{1}{k}\sim \int_{1}^{n}\frac{1}{x}dx .$$

$$ \lim_{n\to \infty }F(n) = \infty. $$

edited Jun 12 '20 at 10:38

Community

1

answered Nov 21 '13 at 07:50

Mhenni Benghorbal

47,431

score 0 · Answer 5 · answered Nov 21 '13 at 08:21

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

Finding the limit $\lim_{n \to \infty}{\frac{\Sigma_{0}^{n}(1/n)}{\ln(n)}}$

5 Answers5

Linked