Results 1 to 2 of 2

Thread: comparison test

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    30

    comparison test

    How would you work out these problems to find out if they converge or diverge?

    Problem 1

    n=1 to infinity 1/(square root (n^3 +2))

    Problem 2

    n=1 to infinity n+2^n/ n^2 2^n

    Problem 3

    n=1 to infinity 3^(n-1) + 1/ 3^n
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, TAG16!

    Converge or diverge?

    $\displaystyle (1)\;\;S \;=\;\sum^{\infty}_{n=1} \frac{1}{\sqrt{n^3+2}} $
    We have: .$\displaystyle \frac{1}{\sqrt{n^3+2}} \:<\:\frac{1}{\sqrt{n^3}} \:=\:\frac{1}{n^{\frac{3}{2}}} $


    $\displaystyle \text{Hence: }\;\sum^{\infty}_{n=1}\frac{1}{\sqrt{n^3+2}} \quad <\; \!\!\! \underbrace{\sum^{\infty}_{n=1}\frac{1}{n^{\frac{3 }{2}}}}_{\text{convergent }p\text{-series}}$

    Therefore, $\displaystyle S$ converges.




    $\displaystyle (2)\;\;S \;=\;\sum^{\infty}_{n=1} \frac{n+2^n}{n^2\!\cdot\!2^n}$

    We have: .$\displaystyle \sum \frac{n+2^n}{n^2\!\cdot\!2^n} \;=\;\sum\left(\frac{n}{n^2\!\cdot\!2^n} + \frac{2^n}{n^2\!\cdot\!2^n}\right) \;=\;\underbrace{\sum\frac{1}{n\!\cdot\!2^n}}_A + \underbrace{\sum\frac{1}{n^2}}_B$

    Series $\displaystyle A\!:\;\;\sum\frac{1}{n\!\cdot\!2^n} \:\leq\:\sum\frac{1}{2^n} $ . . . a convergent geometric series.
    . . Hence, $\displaystyle A$ converges.

    Series $\displaystyle B\!:\;\;\sum\frac{1}{n^2}$ . . . a convergent $\displaystyle p$-series.
    . . Hence, $\displaystyle B$ converges.


    The sum of two convergent series is convergent.

    Therefore, $\displaystyle S$ converges.




    $\displaystyle (3)\;\;S \;=\;\sum^{\infty}_{n=1}\frac{3^{n-1} + 1}{3^n}$

    We have: .$\displaystyle \frac{3^{n-1}+1}{3^n} \;=\;\frac{3^{n-1}}{3^n} + \frac{1}{3^n} \;=\;\frac{1}{3} + \frac{1}{3^n} \;{\color{blue}\;>\;\frac{1}{3}} $

    Hence: .$\displaystyle \sum\frac{3^{n-1}+1}{3^n} \;> \;\sum\frac{1}{3} $ . . . a divergent series

    Therefore, $\displaystyle S$ diverges.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Comparison test and Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Nov 25th 2010, 12:54 AM
  2. Comparison or Limit Comparison Test Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 12th 2010, 07:46 AM
  3. Limit comparison/comparison test series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 25th 2009, 08:27 PM
  4. Comparison & Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 25th 2009, 04:00 PM
  5. Integral Test/Comparison Test
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 27th 2009, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum