Question

Question #1a038

Answer 1

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)`.