Question

Using cauchy's criterion |a_n+p - a_n| 13 years ago

Answer 1

Using \[|a_{n+p}-a _{n}|<\]
Show that \[a_{n}=\frac{ \cos x }{ 3 }+\frac{ \cos 2x }{ 3^{2} } + \frac{ \cos 3x }{ 3^{3} } + .... +\frac{ \cos nx }{ 3^{n} }\] is a cauchy sequence

Answer 2

\[|a_{n+p}-a _{n}|< \epsilon \]

Answer 3

since cosine is bounded above by 1 and below by -1 i believe this is the same as saying
\[\frac{1}{3^{n+p}}-\frac{1}{3^n}|<\epsilon\]

Answer 4

will that be enough ?:D

Answer 5

should be although maybe we need to be a little careful

Answer 6

the largest the absolute value can be is if one cosine is -1 and the other is 1 but the difference is bounded by 2

Answer 7

so you can factor a 2 out of the whole thing, makes no difference in the proof

Answer 8

by which i mean
\[|\frac{\cos((n+p)x)}{a^{n+p}}-\frac{\cos(nx)}{a^n}\leq 2|\frac{1}{3^{n+p}}-\frac{1}{3^n}|\]

Answer 9

\(a\) should be \(3\)