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# How do you test a power series for convergence?

Last updated date: 22nd Feb 2024
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Hint: In order to test the power series for convergence, we can use root test or ratio test, since the convergence of a power series depends on the value of $x$. Therefore, it’s up to us which test we have to use for testing a power series for convergence.

Complete step-by-step solution:
Let us understand with the help of an example.
The interval of convergence of a power series is the set of all x-values for which the power series converges.
Let us find the interval of convergence of:
$\sum\limits_{n=0}^{\infty }{\dfrac{{{x}^{n}}}{n}}$
Now, we will check by Ratio test, so we get:
$\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$
$=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}^{n+1}}}{n+1}.\dfrac{n}{{{x}^{n}}} \right|$
So on further simplifying, we get:
$=\left| x \right|\displaystyle \lim_{n \to \infty }\dfrac{n}{n+1}$
Now, we will apply limit rule as:
$=\left| x \right|.1$
Now, we will apply range to check the power series.
Therefore, we get:
$=\left| x \right|< 1\Rightarrow -1< x < 1$
which means that the power series converges at least on $~\left( -1,1 \right)$.
Now, we need to check its convergence at the endpoints:
$x=-1$ and
$x=1$
If $x=-1$, the power series becomes the alternating harmonic series, so we get:
$\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{n}}$
which is convergent.
So we should include $x=1$:
If $x=1$, the power series becomes the harmonic series, so we get:
$\sum\limits_{n=0}^{\infty }{\dfrac{1}{n}}$
which is divergent. So, $x=1$ should be excluded.
Hence, the interval of convergence is $\left[ -1,1 \right]$

Note: You can think of a power series as a polynomial function of infinite degree since it looks like this:
$\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+...}$
While checking the endpoints of the interval of convergence, they must be checked separately as the Root test and Ratio test are inconclusive here.
To check convergence at the endpoints, we put each endpoint in for $x$, giving us a normal series (no longer a power series) to consider.