Question

arctan summation thing.. 1 sec

Answer 1

\[\sum_{k = 1}^{n} \arctan \left( \frac{ 1 }{ k^{2} + k + 1 } \right)\]

Answer 2

Hmmm, I recall \(\tan^{-1}(x)\) having a very simple MacLaurin series.

Answer 3

Dunno if that would help though, gotta thing...

Answer 4

oh no, i have never worked with those before

Answer 5

if it will help i can research it

Answer 6

Basically do you want an equation in terms of \(n\) then without any summation?

Answer 7

Well is this a math question or a toy question?

Answer 8

oh yes, i am attempting to evaluate the summation the end result should be a term with an n in it

Answer 9

class question I mean

Answer 10

which one is categorized under toy?

Answer 11

Meaning it's not a homework question and you can use any possible method to solve it.

Answer 12

its not a class question. i am looking at some gre stuff and trying to work on the type of math i have not encountered before

Answer 13

gre?

Answer 14

so any ideas are welcome, i am just unsure where to begin

Answer 15

gre = general revised exam, but specifically the math subject exam

Answer 16

Wouldn't you have come across MacLaurin if you're talking a test for graduate students?

Answer 17

depends if you're a math graduate or not

Answer 18

MacLaurin series basically will tell us that: \[ \Large
\tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}
\]

Answer 19

Though I'm not sure this summation will be any easier to be honest.

Answer 20

It is a geometric series at least.

Answer 21

Scratch that, it's an alternating geometric series.

Answer 22

Hmmmmmmm, actually it might just work...

Answer 23

@binarymimic You really wan to learn your series to understand this question.

Answer 24

\[ \large
\begin{split}
\sum_{k = 1}^{n} \arctan \left( \frac{ 1 }{ k^{2} + k + 1 } \right)
&=
\sum_{k = 1}^{n} \sum_{i=0}^{\infty}(-1)^i \left( \frac{ 1 }{ k^{2} + k + 1 } \right)^{2i+1}\frac{1}{2i+1} \\
&=
\sum_{i=0}^{\infty} \sum_{k = 1}^{n} (-1)^i \left( \frac{ 1 }{ k^{2} + k + 1 } \right)^{2i+1}\frac{1}{2i+1} \\
&=
\sum_{i=0}^{\infty}\frac{(-1)^i}{2i+1} \sum_{k = 1}^{n} \left( \frac{ 1 }{ k^{2} + k + 1 } \right)^{2i+1} \\
\end{split}
\]

Answer 25

This is just me messing around.

Answer 26

But it's sort of a dead end now. There is probably some other way.

Answer 27

well i know

\[\arctan \left( \frac{ 1 }{ k^{2} + k + 1 } \right) = \arctan \left( \frac{ (k+1)- k }{ k(k+1) + 1 } \right)\]

if there was some arctan identity that would take the form

\[\arctan \left( \frac{ x - y }{ x*y + 1 } \right)\]

that could help. i'm trying to locate some sort of arctan identity

Answer 28

http://en.wikipedia.org/wiki/List_of_trigonometric_identities says

\[\arctan \alpha - \arctan \beta = \arctan \left( \frac{ \alpha - \beta }{ 1 + \alpha \beta } \right)\]

Answer 29

wait a sec..

Answer 30

@wio

Answer 31

What if you differentiate and then integrate?

Answer 32

@wio what if alpha = k + 1 and beta = k

Answer 33

hmmmm

Answer 34

Then it would be: \[
\tan^{-1}(k+1)-\tan^{-1}(k)
\]?

Answer 35

\[\arctan \left( \frac{ (k + 1) - k }{ 1 + k(k+1) } \right)\]
yeah

Answer 36

then the summation jut telescopes

Answer 37

just*

Answer 38

arctan(k + 1) - arctan(k) + arctan(k+2) - arctan(k + 1) + ... + arcan(n+1)+ arctan(n)

=

-arctan(k) + arctan(n+1) ?

Answer 39

Leaving you with \[
\tan^{-1}(n+1) - \tan^{-1}(1)
\]

Answer 40

well thats silly. are they expecting people to memorize every inverse tangent identity ?

Answer 41

or every inverse trig identity

Answer 42

I don't remember ANY inverse trig identities. I only really use the normal trig ones.

Answer 43

.. i wonder if anyone actually got that question right on a test. anyway thanks a MILLION for your help. i have to read up on taylor and mclaurin series now