# Use the ratio test to find the interval and radius of convergence for each power series. sigma_{n...

## Question:

Use the ratio test to find the interval and radius of convergence for each power series.

{eq}\displaystyle \sum_{n\ =\ 1}^\infty \frac {x^n} {n + 1} {/eq}.

## Ratio Test for Series:

Let {eq}\{a_n\} {/eq} be a sequence of non-zero real numbers. If the limit {eq}\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| {/eq} exists and let {eq}\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|=L {/eq}. Then,

(i) the series {eq}\sum\limits_{n = 1}^\infty {{a_n}} {/eq} converges absolutely if {eq}0 \leq L <1 {/eq} and

(ii) diverges if {eq}L>1{/eq}.

(iii) If {eq}L=1 {/eq} or the limit does not exist then the ratio test is inconclusive.

Alternating Series Test: Assume that {eq}\sum (-1)^n{{a_n}} {/eq} be an infinite sereies, where {eq}a_n\geq 0 {/eq} for all {eq}n\geq 1 {/eq}. If {eq}a_n {/eq} satisfy the following

(i). {eq}\{a_n\} {/eq} is a decreasing sequence

and

(ii). {eq}\mathop {\lim }\limits_{n \to \infty } {a_n}=0 {/eq},

then according to the alternating test, the series {eq}\sum (-1)^n{{a_n}} {/eq} is convergent.

Comparison Test: If the infinite series {eq}\sum_{n=1}^{\infty} b_n {/eq} converges and {eq}0 \le a_n \le b_n {/eq} for all sufficiently large {eq}n {/eq} (that is, for all {eq}n>N {/eq} for some positive integer {eq}N {/eq}), then the infinite series {eq}\sum_{n=1}^{\infty} a_n {/eq} also converges. If the infinite series {eq}\sum_{n=1}^{\infty} b_n {/eq} diverges and {eq}0 \le b_n \le a_n {/eq} for all sufficiently large {eq}n {/eq}, then the infinite series {eq}\sum_{n=1}^{\infty} a_n {/eq} also diverges.

## Answer and Explanation:

Here the given series is {eq}\displaystyle \sum\limits_{n\; = \;1}^\infty {\frac{{{x^n}}}{{n + 1}}} {/eq}.

Let {eq}\displaystyle {a_n} =...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answerHere the given series is {eq}\displaystyle \sum\limits_{n\; = \;1}^\infty {\frac{{{x^n}}}{{n + 1}}} {/eq}.

Let {eq}\displaystyle {a_n} = \frac{1}{{n + 1}}>0,\,\,\,n \ge 1 {/eq}.

Let us calculate the following ratio,

{eq}\displaystyle \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \frac{{\frac{1}{{n + 1 + 1}}}}{{\frac{1}{{n + 1}}}} = \frac{{n + 1}}{{n + 1 + 1}} = \frac{{\frac{{n + 1}}{{n + 1}}}}{{\frac{{n + 1 + 1}}{{n + 1}}}} = \frac{1}{{1 + \frac{1}{{n + 1}}}} {/eq}.

As {eq}\displaystyle n \to \infty {/eq}, we have

{eq}\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{{n + 1}}}} = 1,\,\,\,\,\,\,\,\left( {{\rm{as }}\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 1}} = 0} \right). {/eq}

Clearly, we can see that {eq}\displaystyle \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| =1 {/eq}. Hence the radius of convergence of the series is {eq}\displaystyle 1. {/eq}

Let us check whether the series is convergent at {eq}\displaystyle x=1,-1 {/eq} i.e. at the endpoints.

At {eq}\displaystyle x=1 {/eq}, we have {eq}\displaystyle a_n={a_n} = \frac{1}{{n + 1}}. {/eq}

Then we have {eq}\displaystyle {a_n} = \frac{1}{{n + 1}}>\frac{1}{{n + n}} = \frac{1}{{2n}} = \frac{1}{2}\frac{1}{n} {/eq}.

#### Learn more about this topic:

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 4