# 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.

$\displaystyle \sum_{k=1}^n \frac{1}{k} \approx \ln(\frac{n+e^{1-\gamma}-2}{e^{1-\gamma}}) + \frac{n+1}{n} - \frac{1}{38n} + \frac{1}{9091n}$

Where $\gamma$ is the Euler-Macheroni constant.

$\displaystyle \sum_{k=1}^n \frac{1}{k^2} \approx \frac{591}{400} + \frac{1}{2n^2} + \arccos(1-\frac{(\frac{n-2}{n}-\frac{2}{3})^2+(\frac{1}{n^2}-\frac{1}{36})^2}{8}) - 5.83\times10^{-6}$ for $n\geq 6$ and calculated in radians.

$\displaystyle \sum_{k=1}^n \frac{1}{2k-1} \approx \frac{1}{2}\ln(\frac{2n+2e^{1-\gamma}-5}{2e^{1-\gamma}-1})+1 -0.00531116$

Where $\gamma$ is the Euler-Macheroni constant.

$\displaystyle \sum_{k=1}^n \frac{1}{\sqrt{k}} \approx 2\sqrt{n} + \frac{1}{2\sqrt{n}} - \sqrt{2} - 0.046141$

This is a very good approximation for large $n$ but i have no idea what the constant at the end is. Can anyone help? Note by Chris Sapiano
7 months, 3 weeks ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

- 7 months, 2 weeks ago