Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 4 of 4
  1. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,579
    Thanks
    2,067 times
    Thanked
    659 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #1
    Hey!!

    I want to find for the following series the radius of convergence and the set of $x\in \mathbb{R}$ in which the series converges.

    1. $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{2^n}x^{n^2}}$
    2. $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}}(x-1)^{3n}}$



    I have done the following:

    1. $a_n=\frac{n}{2^n}$
      $$|\frac{a_n}{a_{n+1}}|=\frac{n2^{n+1}}{(n+1)2^n}=\frac{2n}{n+1}=\frac{2}{1+\frac{1}{n}}$$
      So, $$R=\lim_{n\rightarrow \infty}|\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty}\frac{2}{1+\frac{1}{n}}=2$$
      The radius of convergence is $2$. The series converges at $|x|<2$ and diverges at $|x|>2$.
      For $x=\pm 2$ we have the following:
      $$\sum_{n=0}^{\infty}\frac{n}{2^n}(\pm 2)^{n^2}$$
      How could we continue?
    2. $a_n=\frac{1}{(4+(-1)^n)^{3n}}$
      $$\sqrt[n]{|a_n|}=\sqrt[n]{\frac{1}{(4+(-1)^n)^{3n}}}=\frac{1}{(4+(-1)^n)^3}$$
      For $n=2k$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k})^3}=\frac{1}{5^3}$$
      For $n=2k+1$ : $$\lim_{k\rightarrow \infty}\frac{1}{(4+(-1)^{2k+1})^3}=\frac{1}{3^3}$$
      So, $\frac{1}{R}=\lim\sup\sqrt[n]{|a_n|}=\frac{1}{3^3}$.
      The radius of convergence is $3^3=27$. The series converges at $|x|<27$ and diverges at $|x|>27$.
      For $x=27$ we have the following:
      $$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}26^{3n}}$$ How could we check if the series converges?
      For $x=-27$ we have the following:
      $$\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}(-28)^{3n}}=\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}28^{3n}(-1)^n}$$ Is this an alternating series? Do we use here the Leibniz criterium?
    Last edited by mathmari; December 4th, 2016 at 04:39.

  2. MHB Master
    MHB Math Helper

    Status
    Offline
    Join Date
    Jan 2012
    Posts
    1,244
    Thanks
    130 times
    Thanked
    1,991 time
    Thank/Post
    1.600
    #2
    You have the ratio test wrong. It's actually $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ that you need to evaluate.

  3. MHB Master
    MHB Site Helper
    mathmari's Avatar
    Status
    Offline
    Join Date
    Apr 2013
    Posts
    2,579
    Thanks
    2,067 times
    Thanked
    659 times
    Trophies
    1 Highscore
    Awards
    MHB Chat Room Award (2015)  

MHB Model User Award (2015)  

MHB LaTeX Award (2015)
    #3 Thread Author
    Quote Originally Posted by Prove It View Post
    You have the ratio test wrong. It's actually $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ that you need to evaluate.
    Then we would have $\displaystyle \begin{align*} \frac{1}{R}=\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ or not?

  4. MHB Oldtimer
    MHB Site Helper
    MHB Math Scholar
    Opalg's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    Leeds, UK
    Posts
    2,052
    Thanks
    694 times
    Thanked
    5,836 times
    Thank/Post
    2.844
    Awards
    Graduate POTW Award (2016)  

MHB Analysis Award (2016)  

Graduate POTW Award (2015)  

Graduate POTW Award (Jul-Dec 2013)  

MHB Pre-University Math Award (Jul-Dec 2013)
    #4
    Quote Originally Posted by mathmari View Post
    I want to find for the following series the radius of convergence and the set of $x\in \mathbb{R}$ in which the series converges.

    1. $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{2^n}x^{n^2}}$
    2. $\displaystyle{\sum_{n=0}^{\infty}\frac{1}{(4+(-1)^n)^{3n}}(x-1)^{3n}}$
    For a simple power series $\sum a_nx^n$ the radius of convergence is given by $\frac1R = \lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|$ (provided that limit exists). But these are not simple power series, because the power of $x$ is not $n$, but $n^2$ in 1., and $3n$ in 2. In 2., there is the additional complication that the variable is not $x$, but $x-1$.

    So to answer these questions you need to go back to the more general form of the ratio test, which says that a series converges if the limit as $n\to\infty$ of the ratio of the $(n+1)$th term to the $n$th term is less than $1$, and it diverges if that limit is greater than $1$.

    For problem 1., that ratio is $\left|\dfrac{\frac{(n+1)x^{(n+1)^2}}{2^{n+1}}}{\frac{nx^{n^2}}{2^n}}\right| = \Bigl|\dfrac{(n+1)x^{2n+1}}{2n}\Bigr|.$

    For problem 2., you would do best to look at the even-numbered and odd-numbered terms separately. If $n$ is even then the $n$th term is $\Bigl(\dfrac{x-1}5\Bigr)^{3n}.$ If $n$ is odd then it is $\Bigl(\dfrac{x-1}3\Bigr)^{3n}.$
    Last edited by Opalg; December 27th, 2016 at 05:01.

Similar Threads

  1. [SOLVED] Radius of convergence
    By Guest in forum Calculus
    Replies: 3
    Last Post: December 31st, 2015, 01:15
  2. Radius of convergence
    By evinda in forum Analysis
    Replies: 17
    Last Post: June 3rd, 2015, 16:02
  3. radius of convergence
    By brunette15 in forum Calculus
    Replies: 3
    Last Post: June 1st, 2015, 23:34
  4. Radius of Convergence
    By Julio in forum Calculus
    Replies: 2
    Last Post: September 4th, 2014, 11:31
  5. radius of convergence
    By aruwin in forum Calculus
    Replies: 7
    Last Post: July 21st, 2014, 04:21

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards