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)#.