Question

k=1sigma k tends to infinity (1/((k+1)(k+3))) how can i find this series is converges find the closed form for a nth partial sum

Answer 1

converges because denominator is a polynomial of degree 2, numerator is a polynomial of degree 0 and \(2-0=2>1\) so it converges for sure
to add, use partial fractions

Answer 2

can u explain how can i make partial fraction..?

Answer 3

sure
do you know how to do partial fraction decomposition? probably used this for integration

Answer 4

no ..can u explain..

Answer 5

you get
\[\frac{1}{(k+1)(k+3)=\frac{1}}{2}\left(\frac{1}{k+1}-\frac{1}{k+3}\right)\] and then start adding up
you will see that the sum "telescopes" i.e. you will be adding and subtracting the same terms

Answer 6

what is this i can't understnd what are u typing..:(

Answer 7

sorry site was messing up

Answer 8

\[\frac{1}{(k+1)(k+3)}=\frac{1}{2}\left(\frac{1}{k+1}-\frac{1}{k+3}\right)\] is what i was trying to write. now start with \(k=1, 2, 3, ...\) and you will see that the sum telescopes as whatever you add you will subtract later