Sum of Harmonic Series

It is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation.

We need to find the sum of the following series

1a+1a+d+1a+2d++1a+(n1)d\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+\ldots+\dfrac{1}{a+(n-1)d}

Consider the function f(x)=1xf(x)=\frac{1}{x}, we intend to take middle riemann sums with rectangles of width dd starting from x=ax=a to x=a+(n1)dx=a+(n-1)d.

Each rectangle in the figure has a width dd. The height of the ithi\text{th} rectangle is 1a+(i1)d\frac{1}{a+(i-1)d}. The sum of the area of the rectangles is approximately equal to the area under the curve.

Area under f(x) from x=ad2x=a-\frac{d}{2} to x=a+(n12)dn=1nda+(n1)dx=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}

ad2a+(n12)ddxxn=1nda+(n1)d\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}

Let Sn=n=1n1a+(n1)dS_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}

ln2a+(2n1)d2add×Sn\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n

snln2a+(2n1)d2add\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}


Note

  • Apologies for the shabby graph.

  • d0d\neq 0

Note by Aneesh Kundu
4 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

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.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Thanks! This is a very good and useful note.

Akshay Yadav - 4 years, 1 month ago

Log in to reply

Thanks. :)

Aneesh Kundu - 4 years ago

Log in to reply

Hey how you have assigned limit of xx can you please clarify

Atul Shivam - 4 years ago

Log in to reply

xx varies from ad2a-\frac{d}{2} to a+(n12)da+\left(n-\frac{1}{2}\right)d.

Aneesh Kundu - 4 years ago

Log in to reply

yaa ,I got this but also you can't use this formula for finding sum of similar terms

i.e. Sn=12+12+12+12+....12(nthterm)S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term) as common difference is 00 so it will be in indeterminate form

Atul Shivam - 4 years ago

Log in to reply

@Atul Shivam Thanks for the suggestion, I added this point in the note.

Aneesh Kundu - 4 years ago

Log in to reply

@Aneesh Kundu by the way is it real???

Atul Shivam - 4 years ago

Log in to reply

@Atul Shivam This formula gives really good approximations when d0d\rightarrow 0 or for large values of nn with a not so big dd .

Aneesh Kundu - 4 years ago

Log in to reply

@Aneesh Kundu Yes Absolutely

Atul Shivam - 4 years ago

Log in to reply

@Aneesh Kundu Just followed you :-)

Atul Shivam - 4 years ago

Log in to reply

@Aneesh Kundu I mean to is it original(your own)???

Atul Shivam - 4 years ago

Log in to reply

@Atul Shivam Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.

Aneesh Kundu - 4 years ago

Log in to reply

@Aneesh Kundu it's really fantastic

Atul Shivam - 4 years ago

Log in to reply

@Atul Shivam Thanks. :)

Aneesh Kundu - 4 years ago

Log in to reply

your above expression will be incorrect when 2a=d \boxed{2a=d}

Atul Shivam - 4 years ago

Log in to reply

In this case calculate the sum from a2a_2 to ana_n, using the given formula and then add a1a_1 to both sides.

Aneesh Kundu - 4 years ago

Log in to reply

How can area under that curve=d×Sn ??here d is denoted as width

Emran Hossain - 3 years, 9 months ago

Log in to reply

Area under the curve A=1ad+1a+dd+1a+2ddA=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots Ad=(1a+1a+d)\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots ) A=dS˙nA=d\dot S_{n}

Aneesh Kundu - 3 years, 9 months ago

Log in to reply

Ohoo now i understand clearly. Thanks bro...

Emran Hossain - 3 years, 9 months ago

Log in to reply

@Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki.

Akshay Yadav - 3 years, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...