Approximating Sums

Here are my solutions to the approximations of the sums \(\displaystyle \sum_{k=1}^n \frac{1}{k}\) , \(\displaystyle \sum_{k=1}^n \frac{1}{k^2}\), \(\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}}\) and \(\displaystyle \sum_{k=1}^n \frac{1}{2k-1}\) .

I have not researched into whether there is an answer to each of these. I just tried to find them out myself and could not find a solution. Only an approximation. Any comments to refine these or directions to a solution would be appreciated.

∑k=1n1k≈ln⁡(n+e1−γ−2e1−γ)+n+1n−138n+19091n\displaystyle \sum_{k=1}^n \frac{1}{k} \approx \ln(\frac{n+e^{1-\gamma}-2}{e^{1-\gamma}}) + \frac{n+1}{n} - \frac{1}{38n} + \frac{1}{9091n}

Where γ\gamma is the Euler-Macheroni constant.

∑k=1n1k2≈591400+12n2+arccos⁡(1−(n−2n−23)2+(1n2−136)28)−5.83×10−6\displaystyle \sum_{k=1}^n \frac{1}{k^2} \approx \frac{591}{400} + \frac{1}{2n^2} + \arccos(1-\frac{(\frac{n-2}{n}-\frac{2}{3})^2+(\frac{1}{n^2}-\frac{1}{36})^2}{8}) - 5.83\times10^{-6} for n≥6n\geq 6 and calculated in radians.

∑k=1n12k−1≈12ln⁡(2n+2e1−γ−52e1−γ−1)+1−0.00531116\displaystyle \sum_{k=1}^n \frac{1}{2k-1} \approx \frac{1}{2}\ln(\frac{2n+2e^{1-\gamma}-5}{2e^{1-\gamma}-1})+1 -0.00531116

Where γ\gamma is the Euler-Macheroni constant.

∑k=1n1k≈2n+12n−2−0.046141\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}} \approx 2\sqrt{n} + \frac{1}{2\sqrt{n}} - \sqrt{2} - 0.046141

This is a very good approximation for large nn but i have no idea what the constant at the end is. Can anyone help?

#Calculus

Note by Chris Sapiano
1 year, 6 months ago

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Comments

It would be nice to see how you derived these expressions. I can sum the infinite series but this makes me curious

Karan Chatrath - 1 year, 5 months ago
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