Question

How do you test the series `sum_(n=0)^(oo) n/((n+1)(n+2))` for convergence?

Answer 1

The series diverge

Explanation:

Perform the limit comparison test

`a_n=n/((n+1)(n+2))`

and `b_n=1/n`, this series diverge

`a_n>0` and `b_n>0`, `AA n in NN`

`lim_(n->oo)a_n/b_n=lim_(n->oo)((n/((n+1)(n+2)))/(1/n))`

`=lim_(n->oo)(n^2/((n+1)(n+2)))`

`=lim_(n->oo)(n^2/((n^2+3n+2)))`

`=lim_(n->oo)(1/((1+3/n+2/n^2)))`

`=1`

We conclude that, by the limit comparison test that the series `a_n` diverge

Answer 2

The series:

`sum_(n=0)^oo n/((n+1)(n+2))`

is divergent.

Explanation:

The series has only positive terms, so we can use the limit comparison test to compare it with the harmonic series:

`lim_(n->oo) (n/((n+1)(n+2)))/(1/n) = lim_(n->oo) n^2/(n^2+3n+2) = 1`

As the limit is finite and positive the two series have the same character, and we know the harmonic series to be divergent, thus also the series:

`sum_(n=0)^oo n/((n+1)(n+2))`

is divergent.

Answer 3

We can use the integral test to find it diverges.

Explanation:

Using the integral test, we find:

`int x/((x+1)(x+2)) dx = int (2/(x+2) - 1/(x+1)) dx`

`color(white)(int x/((x+1)(x+2)) dx) = 2 ln abs(x+2) - ln abs(x+1) + C`

`color(white)(int x/((x+1)(x+2)) dx) = ln (abs(x+2)^2/abs(x+1)) + C`

`color(white)(int x/((x+1)(x+2)) dx) > ln (abs(x+1)^2/abs(x+1)) + C = ln (abs(x+1)) + C`

`color(white)(int x/((x+1)(x+2)) dx) -> oo" "` as `x -> oo`

So:

`sum_(n=0)^N n/((n+1)(n+2)) -> oo" "` as `N -> oo`

Answer 4

See below.

Explanation:

`n/((n+1)(n+2))=2/(n+2)-1/(n+1)` then

`sum_(n=0)^oo n/((n+1)(n+2)) = 2 sum_(n=0)^oo 1/(n+2) - sum_(n=0)^oo 1/(n+1)`

Now considering

`H = sum_(n=1)^n 1/n` we have

`sum_(n=0)^oo n/((n+1)(n+2)) = 2(H-1)-H = H-2`

but as we know `H` is the so called harmonic series

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

know as divergent.

Resuming

`sum_(n=0)^oo n/((n+1)(n+2))` is divergent.