Solveeit Logo
Login

Question

Question: How do you find the interval of convergence \[\sum {{{( - 1)}^n}\dfrac{{{x^n}}}{n}} \] from \[n = [1...

How do you find the interval of convergence ∑(−1)nxnn\sum {{{( - 1)}^n}\dfrac{{{x^n}}}{n}} from n=[1,∞)n = [1,\infty ) ?

Explanation

Solution

Here in this question we have to find the interval where the function is convergent and the value of n ranges is given. To determine the interval of convergence we use the concept of radius of convergence and hence we determine the solution for the given question.

Complete step by step solution:
If the power series only converges for x=ax = a then the radius of convergence is R=0R = 0 and the interval of convergence is x=ax = a . Likewise, if the power series converges for every x the radius of convergence is R=∞R = \infty and interval of convergence is −∞<x<∞- \infty < x < \infty. Now consider the given power series expansion ∑(−1)nxnn\sum {{{( - 1)}^n}\dfrac{{{x^n}}}{n}}

Given: an=(−1)nxnn{a_n} = {( - 1)^n}\dfrac{{{x^n}}}{n}
The Radius of convergence is defined as R=lim⁡n→∞∣an+1an∣R = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|
Therefore an+1=(−1)n+1xn+1n+1{a_{n + 1}} = {( - 1)^{n + 1}}\dfrac{{{x^{n + 1}}}}{{n + 1}}
∣an+1an∣=∣(−1)n+1xn+1(n+1)(−1)nxnn∣\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{{{( - 1)}^{n + 1}}\dfrac{{{x^{n + 1}}}}{{(n + 1)}}}}{{{{( - 1)}^n}\dfrac{{{x^n}}}{n}}}} \right|
Taking reciprocal the above equation is written as
∣an+1an∣=∣(−1)n+1xn+1(n+1)× n(−1)nxn∣\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{{{( - 1)}^{n + 1}}{x^{n + 1}}}}{{(n + 1)}} \times \dfrac{{\,n}}{{{{( - 1)}^n}{x^n}}}} \right|
On cancelling the terms we have
∣an+1an∣=∣(−1)n(n+1)x∣\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{( - 1)n}}{{(n + 1)}}x} \right|
⇒∣an+1an∣=n(n+1)∣x∣\Rightarrow \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \dfrac{n}{{(n + 1)}}\left| x \right|

Taking the limit as n→∞n \to \infty we have
⇒lim⁡n→∞∣an+1an∣=lim⁡n→∞n(n+1)∣x∣\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{(n + 1)}}\left| x \right|
Applying the limit we get,
lim⁡n→∞∣an+1an∣=∣x∣\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = |x|
Based on the ratio test we have the series is absolutely convergent for ∣x∣<1\left| x \right| < 1 and divergent for ∣x∣>1\left| x \right| > 1. now we have to check for the ∣2x∣=1\left| {2x} \right| = 1.
Suppose if x=1x = 1.
The value of an{a_n} can be written as an=(−1)n(1)nn{a_n} = {( - 1)^n}\dfrac{{{{(1)}^n}}}{n} on simplifying the terms we get an=(−1)nn{a_n} = \dfrac{{{{( - 1)}^n}}}{n}
Therefore we have, the series is a harmonic series and the series is convergent.
Suppose if x=−1x = - 1.
The value of an{a_n} can be written as an=(−1)n(−1)n{a_n} = \dfrac{{{{( - 1)}^n}( - 1)}}{n} on simplifying the terms we get an=(−1)n+1n{a_n} = \dfrac{{{{( - 1)}^{n + 1}}}}{n}. Therefore we have, the series is a harmonic series and the series is divergent.

Therefore the interval of convergence ∑2nxnln⁡ n\sum {\dfrac{{{2^n}{x^n}}}{{\ln \,n}}} is (−1,1]( - 1,1].

Note: To determine the interval of convergence of a power series expansion we use the concept of radius of convergence, by this we can determine the radius and hence on further simplification we determine the interval. To simplify the question we use the simple arithmetic operations.