Hello everyone!!

While researching on a series, I found something magical. Can someone explain this??

Let \[S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ……\]

\[S= (1+\frac{1}{3}+\frac{1}{5}+……)+\frac{1}{2} [1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+……] \]

\[S=(1+\frac{1}{3}+\frac{1}{5}+....)+\frac{S}{2}\]

\[\frac{S}{2}=1+\frac{1}{3}+\frac{1}{5}+…\] \[eq^{n} (i)\]

Now from definition of S, \[\frac{S}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+….\] \[eq^{n} (ii)\]

Comparing \(eq^{n} (i)\) and \(eq^{n} (ii)\) and transposing, We get \[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+…..=0\]

But obviously, \[(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+….>0\]

## Comments

Sort by:

TopNewestHere S is not a converging series. The sum of the given series diverges. So you are treating S as a number, but its sum goes to infinity ( and infinity can not be considered as number , because it does not follow the properties of numbers). So unknowingly you are applying algebraic operations on infinity (for you it is S) , which is a flaw in this your something magical. !!!!!!!! – Sandeep Bhardwaj · 2 years, 6 months ago

Log in to reply

– Pranjal Jain · 2 years, 6 months ago

Thanx dude! I got what you are saying! But I dont know much about convergence of a series! Any good and reliable source? Well, I tried to learn convergence from "Hall and Knight". I need some more help!Log in to reply

– Sandeep Bhardwaj · 2 years, 6 months ago

Go for some UnderGraduate Maths book about sequence and series. You will find the complete conceptual information about the convergence and divergence. Hopefully, it will help you.Log in to reply