Question

How do you find the interval of convergence $\sum{\dfrac{{{4}^{n}}}{{{3}^{n}}+{{5}^{n}}}{{x}^{n}}}$ from $n=[0,\infty )$?

Answer 1

From the question we have been asked to find the interval of convergence. For solving this question first we will use the ratio test. The ratio test formulae for a series like our question will be $\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|$. We will use division operations for further simplification and we solve the given question.

Complete step by step solution:
Firstly, as we mentioned above the ratio test states that a series $\sum\limits_{n=0}^{\infty }{{{a}_{n}}}$ converges absolutely id it obeys the below given condition.
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|<1$
Here we use the basic operation in mathematics which is division for the ratio test.
Let us determine the ratio of the series $\sum{\dfrac{{{4}^{n}}}{{{3}^{n}}+{{5}^{n}}}{{x}^{n}}}$.
$\Rightarrow \left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|=\dfrac{\dfrac{{{4}^{n+1}}}{{{3}^{n+1}}+{{5}^{n+1}}}{{\left| x \right|}^{n+1}}}{\dfrac{{{4}^{n}}}{{{3}^{n}}+{{5}^{n}}}{{\left| x \right|}^{n}}}$
$\Rightarrow 4\left| x \right|\dfrac{{{3}^{n}}+{{5}^{n}}}{{{3}^{n+1}}+{{5}^{n+1}}}$
$\Rightarrow \dfrac{4}{5}\left| x \right|\dfrac{1+{{\left( \dfrac{3}{5} \right)}^{n}}}{1+{{\left( \dfrac{3}{5} \right)}^{n+1}}}$
Now, as $\left( \dfrac{3}{5} \right)<1$ we have that:
$\Rightarrow \displaystyle \lim_{n \to \infty }{{\left( \dfrac{3}{5} \right)}^{n}}=0$
So that:
$\Rightarrow \displaystyle \lim_{n \to \infty }\left( \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right)=\displaystyle \lim_{n \to \infty }\dfrac{4}{5}\left| x \right|\left( \dfrac{1+{{\left( \dfrac{3}{5} \right)}^{n}}}{1+{{\left( \dfrac{3}{5} \right)}^{n+1}}} \right)=\dfrac{4}{5}\left| x \right|$
We can then conclude that for:
$\left| x \right|<\dfrac{5}{4}\Rightarrow \displaystyle \lim_{n \to \infty }\left( \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right)<1$ and the series in the question is absolutely convergent
$\left| x \right|>\dfrac{5}{4}\Rightarrow \displaystyle \lim_{n \to \infty }\left( \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right)>1$ and the given series in the question is divergent
The case where $\left| x \right|=\dfrac{5}{4}$ is indeterminate and we have to analyze in detail.
In the case where $x=\pm \dfrac{5}{4}$ we have:
$\Rightarrow \left| {{a}_{n}} \right|=\left( \dfrac{{{4}^{n}}}{{{3}^{n}}+{{5}^{n}}} \right){{\left( \dfrac{5}{4} \right)}^{n}}=\left( \dfrac{{{5}^{n}}}{{{3}^{n}}+{{5}^{n}}} \right)$
$\Rightarrow \left| {{a}_{n}} \right|=\dfrac{1}{1+{{\left( \dfrac{3}{5} \right)}^{n}}}$
So that:
$\Rightarrow \displaystyle \lim_{n \to \infty }\left| {{a}_{n}} \right|=1>0$
And that means that the given series in the question can converge.
In conclusion the series is convergent in the interval $x\in \left( -\dfrac{5}{4},\dfrac{5}{4} \right)$ where it is also absolutely convergent.

Note: Students must be very careful in doing the calculations. Students should have good knowledge in the concept of limits and continuity as well as its applications like the ratio test. We must know that, the ratio test states that a series $\sum\limits_{n=0}^{\infty }{{{a}_{n}}}$ converges absolutely id it obeys the $\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n}}+1}{{{a}_{n}}} \right|<1$ condition for solving the question.