I have been given:

$$\ln{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\text{ for sufficiently large }n$$

Which I can equate to $\ln(n)=\sum\limits_{i=1}^n \frac{1}{i}$

The series I need to sum is:


I know this can be determined by doing $\ln(80000)-\frac{1}{2}\ln(40000)$, but is it valid to say:

$$\ln{2n-1}=\sum\limits_{i=1}^n \frac{1}{2i-1}=\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\text{ ?}$$

If not, a brief reason why it isn't valid would be appreciated!
Thanks in advance

PS. This is part of a question from the Cambridge maths paper STEP III 2008

  • 1
    $\begingroup$ $Ln(2n-1)$ would be $1+1/2+ ...+ 1/(2n-1)$ $\endgroup$
    – OR.
    Jun 25, 2013 at 16:13
  • $\begingroup$ Your given is not true. The limit of the difference of $\ln n$ and $\sum 1/k$ is a constant, but the constant is not zero. $\endgroup$ Jun 25, 2013 at 16:55

3 Answers 3


$$\sum_{i=1}^n \frac{1}{2 i-1} = \sum_{i=1}^{2 n} \frac{1}{i} - \sum_{i=1}^{n} \frac{1}{2 i} \sim \log{(2 n)} - \frac12 \log{n} = \log{2} + \frac12 \log{n}$$

Meanwhile, in the limit as $n \to \infty$,

$$\log{(2 n-1)} = \log{2} + \log{\left ( n-\frac12\right)} = \log{2} + \log{n} + \log{\left ( 1-\frac{1}{2 n}\right)} \sim \log{2} + \log{n}$$

These approximations simply do not match, even to the lowest order. So, no, they are not the same.


The answer is no. $Ln(3)=1+\frac{1}{2}+\frac{1}{3}$. If your formula were valid, it would be $1+\frac{1}{3}$.

Also, the usual name for these sums is the harmonic numbers, denoted $H_n$.

  • $\begingroup$ Apologies for the slight necro - just wanted to point out the starting line was for "sufficiently large n" - I would doubt many would consider only a couple of terms to be sufficiently large $\endgroup$
    – Dangercrow
    Sep 23, 2015 at 23:31
  • $\begingroup$ $+1$. Nice. It’s a bad idea to use $\ln$ for the Harmonic Number. $\endgroup$ Dec 14, 2020 at 4:25

I think that $\lim_{n\to\infty}\ln (2n-1)$ should be simplified as $\sum^{2n-1}_{i} 1/i$ and not $\sum^{n}_{i} 1/(2i-1)$

So thats why maybe your answer doesn't match.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.