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How to find the summation of the above series upto n=49? I know that there is an Euler's formula for this summation. The formula is ln(n)+gamma, where gamma(Euler's constant)=0.05721 . But as far as I know, the formula is applicable only for large values of n. Will n=49 also work?

DEEPANWITA Roy
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    Harmonic Number Your answer is $H_{50}$. There is not a cleaner way to write this. – JMoravitz Sep 05 '23 at 17:20
  • But how to find the summation without using calculator? – DEEPANWITA Roy Sep 05 '23 at 17:24
  • If you are looking for an exact result... to find the sum without a calculator you need patience... a lot of it, and do each addition manually. If you are okay with an approximation, then $\ln(n)+\gamma$ or $\ln(n)+\gamma+\frac{1}{2n}$ or $\ln(n+\frac{1}{2})+\gamma$ or similar will be fine. You can find $\ln(n)$ using any manner of old techniques from lookup tables, to slide rules, and so on... I would not recommend calculating logarithms by hand otherwise if it can be avoided. – JMoravitz Sep 05 '23 at 17:26

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$$\sum_{n=0}^{p-1} \frac 1{n+1}=H_{p}$$

If $p$ is large $$H_{p}=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12 p^2}+\frac{1}{120 p^4}+O\left(\frac{1}{p^6}\right)$$ that you make shorter transforming the series into a simple Padé approximant $$H_{p}\sim\log (p)+\gamma +\frac{18 p}{36 p^2+6 p+1}$$ whose error is $\frac{13}{2160 p^4}$.

This would give $$H_{50}\sim\log(50)+\gamma+\frac{900}{90301}$$ which is in a absolute error of $9.64\times 10^{-10}$.

Claude Leibovici
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