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
3 years, 10 months ago

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Thanks! This is a very good and useful note.

Akshay Yadav - 3 years, 10 months ago

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Thanks. :)

Aneesh Kundu - 3 years, 10 months ago

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

Atul Shivam - 3 years, 10 months ago

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

Aneesh Kundu - 3 years, 10 months ago

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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 - 3 years, 10 months ago

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

Aneesh Kundu - 3 years, 10 months ago

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

Atul Shivam - 3 years, 10 months ago

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@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 - 3 years, 10 months ago

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@Aneesh Kundu Yes Absolutely

Atul Shivam - 3 years, 10 months ago

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@Aneesh Kundu Just followed you :-)

Atul Shivam - 3 years, 10 months ago

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

Atul Shivam - 3 years, 10 months ago

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@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 - 3 years, 10 months ago

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@Aneesh Kundu it's really fantastic

Atul Shivam - 3 years, 10 months ago

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@Atul Shivam Thanks. :)

Aneesh Kundu - 3 years, 10 months ago

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

Atul Shivam - 3 years, 10 months ago

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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 - 3 years, 10 months ago

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

Emran Hossain - 3 years, 6 months ago

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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, 6 months ago

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Ohoo now i understand clearly. Thanks bro...

Emran Hossain - 3 years, 6 months ago

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@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, 5 months ago

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