Question

what is the convergence of the series [summation from n=1 to infinity of 1/[(n+3)(n+4)]]

Answer 1

convergence of \[\sum_{n=1}^{ \infty} 1/[(n+3)(n+4)]\]

Answer 2

we have 1/(n+3)(n+4)
=((n+4) - (n+3))/(n+3)(n+4)
=1/(n+3) - 1/(n+4)

now on plugging values, we have
1/4 - 1/5 + 1/5 - 1/6 +1/6 - 1/7 ....................
=1/4 --->ans.

Answer 3

use telescoping

Answer 4

how do i then determine the convergence ?

Answer 5

there are bunch of tests ... use one of them
http://en.wikipedia.org/wiki/Convergence_tests

Answer 6

i have tried but am failing to figure out which one to use on this type of series

Answer 7

shubhamsrg already showed it.

Answer 8

best of the test is comparison test ...
1/((n+3)(n+4)) < 1/n^2

and you know 1/n^2 converges from integral test.

Answer 9

for these kinds of stuff ... use integral test.

Answer 10

if you see that kind of series, you must think about how to split the eq

Answer 11

partialising right ?

Answer 12

prove it yourself

Answer 13

if you see that kind of eq, raise the partial fraction, after , when you put the number n =1 ,2,3.... then you can easily cancel the length of the series

Answer 14

am then left with 1\4