Summation of the Inverted Tangents

Geometry Level 4

m = 1 [ tan 1 ( 3 m 2 3 m + 1 m 6 3 m 5 + 3 m 4 m 3 + 1 ) ] = ? \large \sum_{m=1}^{\infty} \left [ \tan^{-1} \left (\frac{3m^2-3m+1}{m^6-3m^5+3m^4-m^3+1} \right ) \right ] = \ ?

This is one of my original problems belonging to the set- Questions I've Made © . Also try its sister problem

π 4 \frac{\pi}{4} π 3 \frac{\pi}{3} π 2 \frac{\pi}{2} π 6 \frac{\pi}{6}

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Sanjeet Raria
Mar 16, 2015

We have ( 3 m 2 3 m + 1 m 6 3 m 5 + 3 m 4 m 3 + 1 ) \large (\frac{3m^2-3m+1}{m^6-3m^5+3m^4-m^3+1}) = ( m 3 m 3 + 3 m 2 3 m + 1 m 6 m 3 3 m 3 ( m 2 m ) + 1 ) \large =(\frac{m^3-m^3+3m^2-3m+1}{m^6-m^3-3m^3(m^2-m)+1}) = ( m 3 [ m 3 1 3 m ( m 1 ) ] 1 + ( m 2 m ) 3 ) \large =(\frac{m^3-[m^3-1-3m(m-1)]}{1+(m^2-m)^3}) = ( m 3 ( m 1 ) 3 1 + ( m ) 3 ( m 1 ) 3 ) \large = (\frac{m^3-(m-1)^3}{1+(m)^3(m-1)^3}) . So that m = 1 tan 1 ( 3 m 2 3 m + 1 m 6 3 m 5 + 3 m 4 m 3 + 1 ) \large \sum_{m=1}^{\infty} \tan^{-1} (\frac{3m^2-3m+1}{m^6-3m^5+3m^4-m^3+1}) = m = 1 tan 1 ( m 3 ( m 1 ) 3 1 + m 3 ( m 1 ) 3 ) \large= \sum_{m=1}^{\infty} \tan^{-1} (\frac{m^3-(m-1)^3}{1+m^3(m-1)^3}) = m = 1 ( t a n 1 m 3 t a n 1 ( m 1 ) 3 ) \large =\sum_{m=1}^{\infty} (tan^{-1}m^3-tan^{-1}(m-1)^3) Telescoping the sum we finally get, = tan 1 ( ) tan 1 0 = π 2 \Large=\tan^{-1}(\infty)-\tan^{-1}0=\boxed{\frac{\pi}{2}}

Other approaches are highly welcomed!!

As you have said "other approaches are highly welcomed!!(sic)" I think if there are any other approach.

Kartik Sharma - 6 years, 2 months ago

Log in to reply

You are never going to know which approach can blow a seemingly 'one way solvable' question. A need for alternative can sometimes be a storehouse of greater creative insights.

Sanjeet Raria - 6 years, 2 months ago
Karan Siwach
Mar 16, 2015

Write the general term of the sum as m = 1 tan 1 ( m 3 ( m 1 ) 3 1 + ( m ) 3 ( m 1 ) 3 ) \large \sum_{m=1}^{\infty} \tan^{-1} (\frac{m^3-(m-1)^3}{1+(m)^3(m-1)^3})

so, now it is clear that it is a telescoping sum.

m = 1 t a n 1 ( m ) 3 t a n 1 ( m 1 ) 3 \large \sum_{m=1}^{\infty} tan^{-1}(m)^3-tan^{-1}(m-1)^3

Hence,

tan 1 ( ) 3 t a n 1 ( 1 1 ) 3 = π / 2 \large \tan^{-1}(\infty)^3-tan^{-1}(1-1)^3 = \pi/2

This is an IIT previous year question.

No its not. They guys can't copy it from me and the chances of them making exactly the same question are low.

Sanjeet Raria - 5 years, 11 months ago

Log in to reply

Actually the problem was not this

The IIT Problem was

n = 0 tan 1 ( 1 n 2 + n + 1 ) \displaystyle{\sum^{\infty}_{n=0}\tan^{-1}(\dfrac{1}{n^2+n+1})} or something like this

Md Zuhair - 4 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...