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.

\(d\neq 0\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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.

Log in to reply

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

Log in to reply

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

Ohoo now i understand clearly. Thanks bro...

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Thanks! This is a very good and useful note.

Log in to reply

Thanks. :)

Log in to reply