Answer
385.5k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)