ElectroSpecter
New member
- Joined
- Jan 29, 2012
- Messages
- 1
Apologies if this is in the wrong forum.
I'm trying to find the rate of convergence of the following sequence:
\(\displaystyle \displaystyle \lim_{n \to \infty} sin{\frac{1}{n}} = 0\).
To begin, I used knowledge of the Maclaurin series for sin (hopefully correctly):
\(\displaystyle \bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\cos{\xi(\frac{1}{n})}}{6}\big(\frac{1}{n}\big)^3\bigg|\).
However, there's already a big discrepancy between this attempt and the actual answer given, which starts:
\(\displaystyle \bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\sin{\xi(n)}}{6}\big(\frac{1}{n}\big)^3\bigg|\).
I'm confused for two reasons: from what I understand, the error term for the series should have a cos in it, not a sin. Furthermore, regardless of whether I'm using cos or sin, shouldn't \(\displaystyle \frac{1}{n}\) be inside the parenthesis, not just \(\displaystyle n\)?
I'm trying to find the rate of convergence of the following sequence:
\(\displaystyle \displaystyle \lim_{n \to \infty} sin{\frac{1}{n}} = 0\).
To begin, I used knowledge of the Maclaurin series for sin (hopefully correctly):
\(\displaystyle \bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\cos{\xi(\frac{1}{n})}}{6}\big(\frac{1}{n}\big)^3\bigg|\).
However, there's already a big discrepancy between this attempt and the actual answer given, which starts:
\(\displaystyle \bigg|\sin{\frac{1}{n}} - 0 \bigg| = \bigg| \frac{1}{n} - \frac{\sin{\xi(n)}}{6}\big(\frac{1}{n}\big)^3\bigg|\).
I'm confused for two reasons: from what I understand, the error term for the series should have a cos in it, not a sin. Furthermore, regardless of whether I'm using cos or sin, shouldn't \(\displaystyle \frac{1}{n}\) be inside the parenthesis, not just \(\displaystyle n\)?