So i have this sequence:

$a_{n}=1+ \frac{1}{3}\cos1 +\frac{1}{3^2}\cos2+ ... +\frac{1}{3^n}\cos(n) $

I have to prove it is convergent, and then calculate the limit. I'm not totally sure how to find the limit of this sequence, so i am stuck at the beginning. Because of the 2nd task, it's probably not the best idea to try to prove it's a Cauchy's sequence, so i guess it's the best to find the limit(at least the candidate) and then prove that it's convergent with that limit by definition. But i am stuck at the begining. Thanks in advance.


Hint: $$ \cos k=\operatorname{Re}\bigl(e^{ik}\bigr) $$ and $$ \sum_{k=0}^n\frac{\cos k}{3^k}=\operatorname{Re}\Bigl(\sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k\Bigr). $$ The sum is the sum of a geometric progression of ratio $e^i/3$. $$ \sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k=\frac{(e^i/3)^{n+1}-1}{e^i/3-1}. $$ Since $|e^i/3|=1/3<1$ we have $$ \lim_{n\to\infty}\sum_{k=0}^n\Bigl(\frac{e^i}{3}\Bigr)^k=\frac{1}{1-e^i/3}=\frac{3}{3-\cos1-i\sin1}. $$ The desired limit is the real part, that is $$ \frac{3(3-\cos1)}{(3-\cos1)^2+\sin^21}. $$

  • $\begingroup$ What does $Re(e^{ik})$ mean? I never saw this before, i'm only at the start of calculus, we didn't use this kind of scripting so. Perhaps an easier concept? $\endgroup$ Jun 29 '15 at 13:11
  • $\begingroup$ Do you know Euler's formula: $$e^{it}=\cos t+i\sin t?$$ I suppose you must know it, since otherwise the problem would be almost unsolvable. $\operatorname{Re}(z)$ means the real part of a complex number $z$. $\endgroup$ Jun 29 '15 at 13:22
  • $\begingroup$ we didn't yet come to that subject(Euler), this is an old exam problem. So i was trying to solve it for extra training. So it is suppose to be solvable. Yeah i know the $Re(z)$ tho. $\endgroup$ Jun 29 '15 at 13:32
  • $\begingroup$ I thought of something now. What if i use that $|\cos(n)| \leq1$ and then i can bound that sequence and find a limit? is that possible? $\endgroup$ Jun 29 '15 at 13:46
  • $\begingroup$ This will give you convergence, but not the value of the sum. $\endgroup$ Jun 29 '15 at 13:56

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