0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $$ \frac{3(3-\cos1)}{(3-\cos1)^2+\sin^21}. $$

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.