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I have to use Cauchy's criterion on $$\sum_{n=1}^\infty \frac{\cos{n}a}{3^n}.$$ Because I'll have the absolute value of it to prove Cauchy and cos n is from [-1,1], so it will actually be from [0,1], can I approximate the fraction to just $\frac{a}{3^n}?$ Thanks!

Stefana
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Notice that $-1 \leq \cos(na) \leq 1$ for any $n$ and $a$. Also, $\sum_{n=i}^{j-1} \frac{1}{3^n} = \frac{3}{2}(3^{-i}-3^{-j})$ is an easy geometric series.

Eric Towers
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  • But isn't it cos(n)*a? so the whole thing would be between -a and a – Stefana Nov 02 '19 at 17:57
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    @StefanaPausan : Cosine (and almost all other functions and operators that are written without parentheses) bind the entire next term -- meaning everything up to the next "$+$", "$-$", "$=$", or the end of an enclosing scope (for instance the end of a numerator or denominator written horizontally, or the end of a radical) . So the argument of cosine here is $na$. – Eric Towers Nov 02 '19 at 18:04
  • Ah, ok. Did not know that. Thank you! – Stefana Nov 02 '19 at 18:06