Question

How do you find the interval of convergence $\sum {\dfrac{{{2^n}{x^n}}}{{\ln \,n}}}$ from $n = [2,\infty )$ ?

Answer 1

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 $x = a$ then the radius of convergence is $R = 0$ and the interval of convergence is $x = a$ . Likewise, if the power series converges for every x the radius of convergence is $R = \infty$ and interval of convergence is $- \infty < x < \infty$. Now consider the given power series expansion $\sum {\dfrac{{{2^n}{x^n}}}{{\ln \,n}}}$.Given: ${a_n} = \dfrac{{{2^n}{x^n}}}{{\ln \,n}}$
The Radius of convergence is defined as $R = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|$
Therefore ${a_{n + 1}} = \dfrac{{{2^{n + 1}}{x^{n + 1}}}}{{\ln \,(n + 1)}}$
$\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{\dfrac{{{2^{n + 1}}{x^{n + 1}}}}{{\ln \,(n + 1)}}}}{{\dfrac{{{2^n}{x^n}}}{{\ln \,n}}}}} \right|$
Taking reciprocal the above equation is written as
$\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{{2^{n + 1}}{x^{n + 1}}}}{{\ln \,(n + 1)}} \times \dfrac{{\ln \,n}}{{{2^n}{x^n}}}} \right|$
On cancelling the terms we have
$\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \left| {\dfrac{{2x\,\ln \,n}}{{\ln \,(n + 1)}}} \right|$
Taking the limit as $n \to \infty$ we have
$\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{2x\,\ln \,n}}{{\ln \,(n + 1)}}} \right|$
We can write 2x before the limit since it is independent to n.
$\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 2x\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\,\ln \,n}}{{\ln \,(n + 1)}}} \right|$
Applying the limit we get
$\mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = 2x$
Based on the ratio test we have the series is absolutely convergent for $\left| {2x} \right| < 1$ and divergent for $\left| {2x} \right| > 1$. Now we have to check for the $\left| {2x} \right| = 1$
Suppose if $x = \dfrac{1}{2}$
The value of ${a_n}$ can be written as ${a_n} = \dfrac{{{2^n}\dfrac{1}{{{2^n}}}}}{{\ln \,n}}$ cancelling the terms we get ${a_n} = \dfrac{1}{{\ln \,n}}$
Therefore we have $\dfrac{1}{{\ln \,n + 1}} > \dfrac{1}{{\ln \,n}}$, the series is harmonic series and the series is divergent.
Suppose if $x = - \dfrac{1}{2}$
The value of ${a_n}$ can be written as ${a_n} = \dfrac{{{2^n}\left( { - \dfrac{1}{{{2^n}}}} \right)}}{{\ln \,n}}$ cancelling the terms we get ${a_n} = \dfrac{{{{( - 1)}^n}}}{{\ln \,n}}$
This is a convergent based on Leibnitz's theorem.
Now taking limit as $n \to \infty$, for the above equation we have
$\mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{\ln \,n}} = 0$ and $\dfrac{1}{{\ln \,n}} > \dfrac{1}{{\ln \,(n + 1)}}$

Therefore the interval of convergence $\sum {\dfrac{{{2^n}{x^n}}}{{\ln \,n}}}$ is $\left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right)$.

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.