Convergence of Complex Sequences at Infinity

[yy205001]

Homework Statement

a) (1+i)^-n as n→∞
b) n/(1+i)ⁿ as n→∞

Homework Equations

The Attempt at a Solution

My answers were divergent for both question because (1+i)ⁿ=sqrt(2)*e^n*pi*i/4, so when n→∞, the limit is varying on the circle with radius sqrt(2). But the solution said both of them equal to 0. How can I get that?

Any help is appreciated.

 

[mfb]

Don't forget the minus sign, and the exponent of the magnitude.

 

[yy205001]

mfb:
But when the minus is there, the value is still varying, it does not go to infinity in denominator for part a. So I still cannot get the limit equal to 0.

 

[HallsofIvy]

My answers were divergent for both question because (1+i)ⁿ=sqrt(2)*e^n*pi*i/4

There's your error (re^{i\theta})^n= r^n e^{ni\theta}. (1+ i)^n= (\sqrt{2})^n e^{n\pi i/4}. You did not take the absolute value to the -n powerr.

, so when n→∞, the limit is varying on the circle with radius sqrt(2). But the solution said both of them equal to 0. How can I get that?

Any help is appreciated.

 

[yy205001]

Hallsoflvy:
oh yeah! So sqrt(2)^-n→0 as n→∞!?

 

[HallsofIvy]

Yes.

 

[yy205001]

Thank you so much!

 

[mfb]

That's what I meant with "the exponent of the magnitude.".