Question

does 2/(n+1) converge/diverge?

Answer 1

neither id say ....

Answer 2

2/(n+1) is just an expression that needs to be applied to something in order to have some sort of convergence or divergence ...

Answer 3

the sequence the is defined by this has a limit that goes to zero; since the bottom goes large the top goes small

Answer 4

not the top perse; but the value :)

Answer 5

welll there is the summation, I didn't put it in. \[\sum_{n = 0}^{\infty}\]2/(n+1)

Answer 6

the partial sums .... i cant tell at the moment tho

Answer 7

is that for the sequence of the partial sums?

Answer 8

which test should I use?

Answer 9

no, it is not for partial sums.

Answer 10

then id just play it out as is; cant recall any details for the "tests" at the moment.

the limit of 2/(n+1) as n approaches infinity is 0 since the bottom outweighs the top for large values of n

Answer 11

the sequence is really a set of the range of the function. So whatever the function behaves for large values of n; it either goes to zero and converges, or it doesnt and diverges ...

Answer 12

ok, thank you

Answer 13

youre welcome :)

Answer 14

Your summation is
\[\sum_{n=0}^{\infty}\frac{2}{n+1}= 2\sum_{n=1}^{\infty}\frac{1}{n}\]
which is 2 times the harmonic series. This series is famous for not converging. You can show this using the integral test.