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

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

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

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

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

\[\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}\]

Let \(S_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}\)

\[\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n\]

\[\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}\]

**Note**

Apologies for the shabby graph.

\(d\neq 0\)

## Comments

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TopNewest@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 · 1 year, 4 months ago

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How can area under that curve=d×Sn ??here d is denoted as width – Emran Hossain · 1 year, 5 months ago

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– Aneesh Kundu · 1 year, 5 months ago

Area under the curve \[A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots \] \[\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots ) \] \[A=d\dot S_{n}\]Log in to reply

– Emran Hossain · 1 year, 5 months ago

Ohoo now i understand clearly. Thanks bro...Log in to reply

your above expression will be incorrect when \[ \boxed{2a=d}\] – Atul Shivam · 1 year, 9 months ago

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– Aneesh Kundu · 1 year, 9 months ago

In this case calculate the sum from \(a_2\) to \(a_n\), using the given formula and then add \(a_1\) to both sides.Log in to reply

Hey how you have assigned limit of \(x\) can you please clarify – Atul Shivam · 1 year, 9 months ago

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– Aneesh Kundu · 1 year, 9 months ago

\(x\) varies from \(a-\frac{d}{2}\) to \(a+\left(n-\frac{1}{2}\right)d\).Log in to reply

i.e. \(S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term)\) as common difference is \(0\) so it will be in indeterminate form – Atul Shivam · 1 year, 9 months ago

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– Aneesh Kundu · 1 year, 9 months ago

Thanks for the suggestion, I added this point in the note.Log in to reply

– Atul Shivam · 1 year, 9 months ago

I mean to is it original(your own)???Log in to reply

– Aneesh Kundu · 1 year, 9 months ago

Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.Log in to reply

– Atul Shivam · 1 year, 9 months ago

it's really fantasticLog in to reply

– Aneesh Kundu · 1 year, 9 months ago

Thanks. :)Log in to reply

– Atul Shivam · 1 year, 9 months ago

by the way is it real???Log in to reply

– Aneesh Kundu · 1 year, 9 months ago

This formula gives really good approximations when \(d\rightarrow 0\) or for large values of \(n\) with a not so big \(d\) .Log in to reply

– Atul Shivam · 1 year, 9 months ago

Just followed you :-)Log in to reply

– Atul Shivam · 1 year, 9 months ago

Yes AbsolutelyLog in to reply

Thanks! This is a very good and useful note. – Akshay Yadav · 1 year, 9 months ago

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– Aneesh Kundu · 1 year, 9 months ago

Thanks. :)Log in to reply