Question #1a038

1 Answer
Nov 11, 2017

The interval of convergence is [-1,1).

Explanation:

First, use the Ratio Test to find where the ratio is less than one. Then, after that, check your endpoints.

Let a_n={x^n}/n. The n+1 term is then a_(n+1)=(x^{n+1})/(n+1).

Use the Ratio Test.

lim_{n to infty}|{a_{n+1}}/{a_n}| =lim_(n to infty)|(x^(n+1))/(n+1)cdot(n)/(x^n)|

Cancel out common factors.

=lim_(n to infty)|(xn)/(n+1)|

Pull abs(x) out of the limit.

=abs(x)lim_(n to infty)|(n)/(n+1)|

Multiply the numerator and the denominator by 1/n.

=abs(x)lim_(n to infty)|(1)/(1+1/n)|=abs(x)

The ratio is less than one, whenever

|x|<1 or -1 < x < 1

So, the power series converges at least on the open interval (-1,1).

Now, we need to check the endpoints: x=-1 and x=1.

When x=-1, the series becomes

sum_{n=1}^infty((-1)^n}/n.

We can prove that the series is convergent using the Leibniz theorem as:

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

and:

1/n > 1/(n+1)

When x=1, the series becomes

sum_{n=1}^infty 1/n .

and we know that the harmonic series is divergent based on the p-series test.

Hence, the interval of convergence is [-1,1).