My Math Forum nth - root test - is this a mistake?? Help

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 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 nth-root 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: 436 Thanks: 147 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: 436 Thanks: 147 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: 436 Thanks: 147 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: 7,842 Thanks: 1068 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

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