Question

Sum of this complicated series, should i use some representation before finding its sum?

Answer 1

qn 29

Answer 2

use a riemenn sum

Answer 3

but remember to set your limits from negative infinity to zero before summing

Answer 4

but only use complex numbers

Answer 5

riemann? is there another way ?

Answer 6

remember you are operating in 4-space which you cannot imagine

Answer 7

It's a telescoping series so you really don't have to worry about much.

Answer 8

Just write out the first 2 terms and you'll notice what's gonna happen. Promise. =P

Answer 9

#worstadviceever

Answer 10

#me

Answer 11

hmm hold on,

Answer 12

lol @meshlogic

Answer 13

okay, first few terms hmm.. arctan(2)-arctan(1)+arctan(3)-arctan(2)+arctan(4)-arctan(3)

Answer 14

only -arctan(1) remains.

Answer 15

how do u know if we needed to use telescoping series? if the terms in the function adds/subtracts then use telescoping?

Answer 16

What do you mean? A telescoping series is really just saying that your sum is really a combination of two separate sums that are identical but opposite and shifted by a few indices.

So we could have written the sum as:
\[\sum_{n=1}^{\infty}[\tan^{-1}(n+1)-\tan^{-1}n]=\sum_{n=1}^{\infty}[\tan^{-1}(n+1)]-\sum_{n=1}^{\infty}[\tan^{-1}n]\] replacing n+1 with k, in math terms, n+1=k we have n=k-1 so we plug that into the index of the first one. \[\sum_{k-1=1}^{\infty}[\tan^{-1}k]=\sum_{k=2}^{\infty}[\tan^{-1}k]=\sum_{n=2}^{\infty}[\tan^{-1}n]\] since k and n are just dummy variables, they don't matter so now you can see that you really have the same sum but you started at 2 instead of 1.

So if you just take one term out of the other sum you get: \[-\tan^{-1}(1)+\sum_{n=2}^{\infty}[\tan^{-1}n]-\sum_{n=2}^{\infty}[\tan^{-1}n]=-\tan^{-1}(1)+0\]

Answer 17

ooo thiis!! thank you! my doubts are cleared! i always used to just expand telescoping series, but this is a brilliant method! :D

Answer 18

Yes, This is called method of difference, it is taught alevel further math.