# Ln Series Summation

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:

$$\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{79999}$$

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!

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

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$Ln(2n-1)$ would be $1+1/2+ ...+ 1/(2n-1)$ – Mlazhinka Shung Gronzalez LeWy Jun 25 '13 at 16:13
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. – Thomas Andrews Jun 25 '13 at 16:55

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

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.

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

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 $H_n$.