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# 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

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

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.

• $$d\neq 0$$

Note by Aneesh Kundu
1 year 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. · 7 months, 2 weeks ago

How can area under that curve=d×Sn ??here d is denoted as width · 8 months, 2 weeks ago

Area under the curve $A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots$ $\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$ · 8 months, 2 weeks ago

Ohoo now i understand clearly. Thanks bro... · 8 months, 2 weeks ago

your above expression will be incorrect when $\boxed{2a=d}$ · 1 year 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. · 1 year ago

Hey how you have assigned limit of $$x$$ can you please clarify · 1 year ago

$$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left(n-\frac{1}{2}\right)d$$. · 1 year ago

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

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 · 1 year ago

Thanks for the suggestion, I added this point in the note. · 1 year ago

I mean to is it original(your own)??? · 1 year 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. · 1 year ago

it's really fantastic · 1 year ago

Thanks. :) · 1 year ago

by the way is it real??? · 1 year ago

This formula gives really good approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ . · 1 year ago

Just followed you :-) · 1 year ago

Yes Absolutely · 1 year ago

Thanks! This is a very good and useful note. · 1 year ago