Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The function is $$ f(n) = \sum_{i=1}^{n} \frac{1}{2i-1}$$ How can I compute for example $f(20)$ or $f(50)$ without using a calculator. I want to have an approximation

share|cite|improve this question
are you allowed to use a table of log-values? –  Alex May 24 '14 at 9:29
yeah, log values are ok –  neticin May 24 '14 at 9:45
pls see the edit –  Alex May 24 '14 at 9:57

2 Answers 2

up vote 5 down vote accepted

It's a harmonic sum with odd terms: $$ S_n = 1 + \frac{1}{3} + \ldots \frac{1}{2n-1} = 1 +\frac{1}{2}+\frac{1}{3} + \ldots \frac{1}{2n} - \frac{1}{2}-\frac{1}{4} -\ldots \frac{1}{2n} = H_{2n}-\frac{1}{2}H_n $$ and each $H_{k} \sim \log k +\gamma$.

EDIT: Since $H_{2n} \sim \log 2n + \gamma = \log 2 + \log n + \gamma$ this whole expression reduces to $\frac{1}{2} \log n +\log 2 +\frac{\gamma}{2}$

share|cite|improve this answer

It is possible to show that $$f(n) = \sum_{i=1}^{n} \frac{1}{2i-1}=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{1}{2}\right)-\psi ^{(0)}\left(\frac{1}{2}\right)\right)$$ where appears the polygamma function. For large values of $n$, you have as an approximation $$f(n) =\left(-\frac{1}{2} \log \left(\frac{1}{n}\right)-\frac{\psi ^{(0)}\left(\frac{1}{2}\right)}{2}\right)+O\left(\left(\frac{1}{n}\right)^2\right)$$ where $$\psi ^{(0)}\left(\frac{1}{2}\right) \simeq -1.96351$$ So $$f(n) \simeq \frac{1}{2} \log(n)+0.981755$$ will be a good approximation.

Let us try with you numbers : for $n=20$, the exact value is $2.479673210$ while the approxiamtion leads to $2.479621150$; for $n=50$, the corresponding values are $2.937774848$ and $2.937766516$.

share|cite|improve this answer
how did you get digamma of 1/2 without a calculator or a table? –  Alex May 24 '14 at 9:45
As far as I understood, the OP wanted an approximation. I did not think that he required that the approximation should not be built using a computer. By the way, I like your answer : +1 ! Cheers. –  Claude Leibovici May 24 '14 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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