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)=x1, 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 ith rectangle is a+(i−1)d1. The sum of the area of the rectangles is approximately equal to the area under the curve.
Area under f(x) from x=a−2d to x=a+(n−21)d≈n=1∑na+(n−1)dd
⇒∫a−2da+(n−21)dxdx≈n=1∑na+(n−1)dd
Let Sn=n=1∑na+(n−1)d1
ln2a−d2a+(2n−1)d≈d×Sn
⇒sn≈dln2a−d2a+(2n−1)d
Note
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Comments
Thanks! This is a very good and useful note.
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Thanks. :)
Hey how you have assigned limit of x can you please clarify
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x varies from a−2d to a+(n−21)d.
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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=21+21+21+21+....21(nthterm) as common difference is 0 so it will be in indeterminate form
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d→0 or for large values of n with a not so big d .
This formula gives really good approximations whenLog in to reply
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your above expression will be incorrect when 2a=d
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In this case calculate the sum from a2 to an, using the given formula and then add a1 to both sides.
How can area under that curve=d×Sn ??here d is denoted as width
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Area under the curve A=a1d+a+d1d+a+2d1d… dA=(a1+a+d1…) A=dS˙n
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Ohoo now i understand clearly. Thanks bro...
@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.