Question

How do you find the interval of convergence `Sigma ((-1)^n(x+2)^n)/n` from `n=[1,oo)`?

Answer 1

The series:

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

is absolutely convergent for `x in (-3,-1)` and simply convergent for `x =-1`

Explanation:

To find the interval of convergence we can apply the ratio test, stating that a necessary condition for a series `sum_(n=1)^oo a_n` to converge is that:

`L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1`

If `L < 1` the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

`abs (a_(n+1)/a_n) = abs (frac ( (x+2)^(n+1)/(n+1)) ((x+2)^n/n)) = abs(x+2) n/(n+1)`

so that:

`lim_(n->oo) abs (a_(n+1)/a_n) = abs(x+2)`

we can therefore conclude that the series is absolutely convergent for `abs (x+2) < 1` and divergent for `abs (x+2) > 1`.

In the case where `abs (x+2) = 1` we know that the test is inconclusive but we can see that for `x=-3` the series becomes:

`sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n`

that is divergent, while for `x= -1` we have:

`sum_(n=1)^oo (-1)^n/n`

that is convergent, but not absolutely convergent.

In conclusion the series converges for `x in (-3,-1]`