Question

What is the interval of convergence of `sum (x+2)^n / sqrt(n)`?

Answer 1

The series:

`sum(x+2)^n/sqrtn`

is convergent in the interval `x in [-3,-1)` and absolutely convergent in the interior of the interval.

Explanation:

For `x in (-3,-1)` we have that `abs(x+2) < 1` so that:

`abs(x+2)^n/sqrtn <= abs(x+2)^n`

and as:

`sum abs(x+2)^n`

is convergent because it is a geometric series of ratio `< 1` then also our series is absolutely convergent by direct comparison.

For `x in [-1,+oo)` we have that `(x+2) >= 1` so that:

`(x+2)^n/sqrtn >= 1/sqrtn`

and as:

`sum 1/sqrtn`

is divergent based on the `p-`series test, then also our series is divergent.

For `x in (-oo, -3)` we have that `(x+2) <0` so that:

`sum (x+2)^n/sqrtn`

is an alternating series.
To simplify notation we pose: `a=abs(x+2) > 1` and note that, if:

`(1) lim_(t->oo) a^t/sqrtt`

exists, then we must have:

`lim_(n->oo) a^n/sqrtn = lim_(t->oo) a^t/sqrtt`

The limit (1) is in the form `oo/oo` so we can solve it using l'Hospital's rule:

`lim_(t->oo) a^t/sqrtt = lim_(t->oo) (d/dt a^t)/( d/dt sqrtt)=lim_(t->oo) (a^tlna)/( 1/(2sqrtt))=lim_(t->oo) 2lnasqrtta^t=+oo`

Then we know that:

`lim_(n->oo) (x+2)^n/sqrtn != 0`

and the series is not convergent.

Finally, for `x=-3` we have that the series becomes:

`sum (-1)^n/sqrtn`

this is also an alternating series and we can apply Leibniz test. CLearly:

`lim_(n->oo) 1/sqrtn =0`

and

`1/sqrt(n+1) < 1/sqrtn` for `n>=1`

then Leibniz test is satisfied and the series is convergent.

Answer 2

`-3lt=xlt-1`

Explanation:

This question assumes the infinite sum `sum_(n=1)^oo(x+2)^n/sqrtn`.

For the given series `suma_n`, find the ratio `a_(n+1)/a_n`:

`a_(n+1)/a_n=((x+2)^(n+1)/sqrt(n+1))/(((x+2)^n)/(sqrtn))=(x+2)^(n+1)/(x+2)^n(sqrtn/sqrt(n+1))=(x+2)sqrt(n/(n+1))`

Now, find `lim_(nrarroo)abs(a_(n+1)/a_n)`:

`lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs((x+2)sqrt(n/(n+1)))`

The limit only depends on how `n` changes, so the term that depends only on `x` can be extracted from the limit:

`=abs(x+2)lim_(nrarroo)abssqrt(n/(n+1))`

This limit approaches `1`:

`=abs(x+2)`

The ratio test states that for the series `suma_n`, the series converges when `lim_(nrarroo)abs(a_(n+1)/a_n)<1`. So, the given series converges when:

`abs(x+2)<1`

Or, dealing away with the absolute value:

`-1ltx+2lt1`

So:

`-3ltxlt-1`

Before we call this our interval of convergence, we must test the endpoints `x=-3` and `x=-1` in the original series. These endpoints may either converge or diverge, and we need to see which act which way.

At `x=-3`, the series becomes:

`sum_(n=1)^oo(-3+2)^n/sqrtn=sum_(n=1)^oo(-1)^n/sqrtn`

Although the series `sum_(n=1)^oo1/sqrtn` would diverge through the `p`-test, since `p=1//2`, this is an alternating series by virtue of the `(-1)^n`. Since `lim_(nrarroo)1//sqrtn=0` and `sqrtn>0`, this is a (conditionally) convergent series through the alternating series test.

Thus, `x=-3` is included in the interval of convergence.

Test the other endpoint, `x=-1`:

`sum_(n=1)^oo(-1+2)^n/sqrtn=sum_(n=1)^oo1/sqrtn`

which diverges. Thus, `x=-1` is not included in the interval of convergence.

This leads to the interval:

`-3lt=xlt-1`