Question

Prove that this series converges or diverges, and compute its sum on a) :

a) Sum {arctg[1/(n^2+n+1)]}
b) Sum {sqrt(2n-1)ln(4n+1)/n(n+1)}

Answer 1

from n=0 or from n=1 ?

Answer 2

\[\sum_{n=0}^\infty\tan^{-1}(\frac1{n^2+n+1})\]?

Answer 3

ups, a) from n=1 to infinity and b) from n=2 to..

Answer 4

\[\sum_{n=1}^\infty\tan^{-1}(\frac1{n^2+n+1})\]

Answer 5

yep

Answer 6

the book suggest tg(a-b)=[tg(a)-tg(b)]/[1+tg(a)tg(b)]

Answer 7

would be useful for a)

Answer 8

I thought cause book's suggestion to transform the general term in some way to related it with telescopic series, but I don't

Answer 9

it's inverse tangent though in the problem, right?

Answer 10

yep

Answer 11

@ash2326 @KingGeorge @myininaya series help?

Answer 12

what about if i put the general term like

Answer 13

1/n(n+1)+1=(n-n+1)/n(n+1)+1 and do the changes n=tg(a) and n+1=tg(b), use the above formula for the tangent and turn into telescopic series? Just an idea, I'm lose...xd

Answer 14

*lost

Answer 15

It is well known that
\[
\tan ^{-1}(x)+\tan ^{-1}(y)=\tan
^{-1}\left(\frac{x+y}{1-x
y}\right)\\
x=-n\\
y= n+1\\
\tan ^{-1}(n+1)-\tan ^{-1}(n)=\tan
^{-1}\left(\frac{1}{n^2+n+1}\right)\\
Let c_n=\tan^{-1} (n)\\
then\\
a_n=\tan
^{-1}\left(\frac{1}{n^2+n+1}\right)= c_{n+1} - c_n\\

\]
Use telescoping to get
\[
s_n=c_{n+1}- c_1=c_{n+1}- \frac \pi 4
\]

Answer 16

\[
\lim_{n\to \infty} c_{n+1} =\frac \pi 2\\
\lim_{n\to \infty} s_n= \frac \pi 2 - \frac \pi 4=\frac \pi 4
\]

Answer 17

So the sum of the first integral is \(\frac \pi 4\)

Answer 18

Did you understand it?

Answer 19

yes, thanks a lot!

Answer 20

and what about b) comparison test? limit test?

Answer 21

b) solved, by limit test, if anybody wants a detailed explanation, just ask