
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
May 1st, 2018, 09:27 PM  #1 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0  nth  root test  is this a mistake?? Help
I need help with the following, but I am stuck. Is the i to the power n a typo? Please help me solve this question... "Use the nthroot test to find the conditions on the constant a in order that the series $\displaystyle \sum _{n=1}^{\infty \:}\left(\frac{an}{an+1}\right)^{n^2}i^n$ is absolutely convergent Last edited by skipjack; May 2nd, 2018 at 02:37 PM. 
May 1st, 2018, 10:45 PM  #2 
Senior Member Joined: Oct 2009 Posts: 555 Thanks: 179 
We can't possibly know. You'll need to ask the person who gave you this problem.

May 1st, 2018, 11:12 PM  #3 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
Thanks... i meant i have no idea how one would go about solving it. Can you perhaps assist?

May 1st, 2018, 11:23 PM  #4 
Senior Member Joined: Oct 2009 Posts: 555 Thanks: 179 
Sure, can you tell me what $a$ is and what $i$ is? Can we assume $i$ is the imaginary unit? Is $a$ complex? Real?

May 1st, 2018, 11:36 PM  #5 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
I made a mistake  it should be the constant α not a i is the imaginary unit yes 
May 2nd, 2018, 02:31 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Well, then, do you know what the "nth root test" (I believe I would just say "root test") is? If you do then what do you get when you apply it?

May 2nd, 2018, 04:08 AM  #7 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
Yes, here is what I have so far $\displaystyle \:\lim _{n\to \infty }\:\sqrt[n]{\left(\frac{αn}{αn+1}\right)^{n^2}i^n}$ which, after simplifying, leaves me with $\displaystyle \:\lim _{n\to \infty }\:\left(\frac{αn}{αn+1}\right)^ni$ To simplify I though of multiplying with $\displaystyle \frac{\frac{1}{αn}}{\frac{1}{αn}}$ which is equal to 1 Then I would be left with $\displaystyle \lim _{n\to \infty }\left(\frac{1}{1+\frac{1}{n}}\right)^n\:i$ So then because $\displaystyle \lim _{n\to \infty }\left(1+\frac{1}{n}\right)^n\:\:=\:e$ I am left with $\displaystyle \lim _{n\to \infty }\left(\frac{1}{e}\right)i$ I don't know if I am allowed doing that and I don't know how to go further to "find the conditions on constant $\displaystyle α$" in order that the series is absolutely convergent'' 
May 2nd, 2018, 04:14 AM  #8 
Senior Member Joined: Oct 2009 Posts: 555 Thanks: 179 
You take the nth root of the modulus in the nth root test. You forgot the modulus.

May 2nd, 2018, 05:33 AM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 
What happened to $\alpha$?

May 2nd, 2018, 07:12 AM  #10 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
so will it be right to take the complex number as $\displaystyle 0+\left(\frac{1}{e}\right)i$ and then use the formula $\displaystyle \sqrt{a^2+b^2}$ ? Please help... I'm more lost now than before 

Tags 
mistake, nth, root, test 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Nth Root Test  mahowm10  Calculus  2  December 5th, 2014 12:15 PM 
Can someone fix my mistake  livestrong136  Calculus  1  May 8th, 2012 04:02 AM 
Is there any mistake  bobar77  Algebra  8  February 17th, 2012 09:37 PM 
where is the mistake??  islam  Calculus  6  December 3rd, 2010 07:33 PM 
Root Test  sis3  Real Analysis  1  May 3rd, 2010 01:16 PM 