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Answer 1 · 2017-11-11T01:54:56

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

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

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

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 $[-1,1)$ .