# Summation of tan inverse 1/(2x^2) from 1 to infinity

JUSTANOTHERLOSER
28 Points
5 years ago
It`s a bit tricky question. Your can solve it like this : We have to find sum of arctan(1/2n^2) n→∞. First let us find this for a finite vale of `n` and then for n→∞. Step 1: 1/2n^2 = 2/4n^2 = ((2n+1)-(2n-1))/1+(2n+1)(2n-1)Step 2:Here we use the identityArctan(a) - arctan(b) = arctan[(a-b)/1+ab]Let a =2n+1, b=2n-1Then we getArctan(2n+1)-arctan(2n-1) = arctan[(2n+1)-(2n-1))/1+(2n+1)(2n-1)]But from Step 1, we get the following Arctan(2n+1)-arctan(2n-1) = arctan(1/2n^2)Step 3:Now we perform the summation.Let each term of the summation. Let the sum be S. S= arctan[1/2(1^2)] + arctan[1/2(2^2)] + arctan[1/2(3^2)] +...........................................+arctan[1/2(n-1)^2] + arctan[1/2n^2]Now we apply the results from Step2.S = [arctan(3) - arctan(1)] + [arctan(5) - arctan(3)] + [arctan(7) - arctan(5)] + ....................+ [arctan(2(n-1)+1) - arctan(2(n-1)-1)] + [arctan(2n+1) - arctan(2n-1)]Now if you observe carefully, you will notice that all the terms in the sum S cancel out except arctan(2n+1) and -arctan(1). Applying this we get S= arctan(2n+1) - arctan(1)Again applying the identityArctan(a) - arctan(b) = arctan[(a-b)/1+ab]We get,S= arctan[((2n+1)-1)/1+(2n+1)(1)]S= arctan[2n/(2n+2)]We get the final summation as S= arctan[n/(n+1)]Step4:For finding the sum S when n→∞We simply do the following:S= arctan[n/(n+1)]Now we write (n+1)=n(1+1/n)S= arctan[n/n(1+(1/n))]We get,S=arctan[1/(1+1/n)]Now applying the limit,limit S = lim arctan[1/(1+1/n)]n→∞ n→∞But as n→∞, 1/n → 0. Using this we get,limit S = arctan[1/(1+0)] = arctan(1) = π/4n→∞So the final answer is π/4
one year ago
Dear student,

It`s a bit tricky question. Your can solve it like this : We have to find sum of arctan(1/2n^2) n→∞.
First let us find this for a finite value of `n` and then for n→∞.
Step 1: 1/2n^2 = 2/4n^2 = ((2n+1)-(2n-1))/1+(2n+1)(2n-1)
Step 2:Here we use the identityArctan(a) - arctan(b) = arctan[(a-b)/1+ab]
Let a =2n+1, b=2n-1
Then we getArctan(2n+1)-arctan(2n-1) = arctan[(2n+1)-(2n-1))/1+(2n+1)(2n-1)]
But from Step 1, we get the following
Arctan(2n+1)-arctan(2n-1) = arctan(1/2n^2)
Step 3:Now we perform the summation.Let each term of the summation.
Let the sum be S. S= arctan[1/2(1^2)] + arctan[1/2(2^2)] + arctan[1/2(3^2)] +...........................................+arctan[1/2(n-1)^2] + arctan[1/2n^2]
Now we apply the results from Step2.S = [arctan(3) - arctan(1)] + [arctan(5) - arctan(3)] + [arctan(7) - arctan(5)] + ....................+ [arctan(2(n-1)+1) - arctan(2(n-1)-1)] + [arctan(2n+1) - arctan(2n-1)]
Now if you observe carefully, you will notice that all the terms in the sum S cancel out except arctan(2n+1) and -arctan(1). Applying this we get S= arctan(2n+1) - arctan(1)
Again applying the identity
Arctan(a) - arctan(b) = arctan[(a-b)/1+ab]
We get,S= arctan[((2n+1)-1)/1+(2n+1)(1)]
S= arctan[2n/(2n+2)]
We get the final summation as
S= arctan[n/(n+1)]
Step4:For finding the sum S when n→∞
We simply do the following:
S= arctan[n/(n+1)]
Now we write
(n+1)=n(1+1/n)S= arctan[n/n(1+(1/n))]
We get, S=arctan[1/(1+1/n)]
Now applying the limit,limit S = lim arctan[1/(1+1/n)]n→∞ n→∞
But as n→∞, 1/n → 0.
Using this we get,limit S = arctan[1/(1+0)] = arctan(1) = π/4n→∞
So the final answer is π/4

Thanks and regards,
Kushagra