Question

Calculus II, Sequence Convergence and Divergence: I don't understand how to evaluate this sequence as converging or diverging, posted below in a minute.

Answer 1

\[a _{n} = \frac{ (-1)^{n +1} }{ 2n -1 }\]
That's all well and good, but when taking the limit of it as n approaches infinity, I get the numerator resulting in a DNE and the denominator resulting in the whole thing effectively equaling zero.

Answer 2

This is where I get confused; there's got to be some sort of algebraic manipulation to remove that quality of the numerator being DNE, but I don't know what to do in order to get there, or it's at least not readily apparent.

Answer 3

\[\lim_{n \rightarrow \infty} \frac{ (-1)^{n+1} }{ 2n-1 } = \frac{ (-1)^{\infty} }{ \infty } \]This looks like it's almost suitable for l'Hopital's rule, but it's not.

Answer 4

(The answer is that it converges to zero.)

Answer 5

use the absolute convergence to effectively ignore the alternating signs

Answer 6

I have no idea what that is and don't think I've studied it yet, so I'm not quite sure how-or if I'm allowed to-utilize it. Would there be any way to reach the same conclusion using other Calculus II/prior math?

Answer 7

(If it's just as straightforward as the Wikipedia article on it seems - http://en.wikipedia.org/wiki/Absolute_convergence ), I'm a little confused because literally like a page or two ago, it emphasized the fact that \[\lim_{n \rightarrow \infty}(-1)^{n}\]did not exist, and therefore concluded that the series diverged.

Answer 8

On the other hand, you can also use the Squeeze Theorem on something like this:
\[\frac{-1}{2n-1}\le a_n\leq\frac{1}{2n-1}\]

Answer 9

\[(-1)^n\]
does not converge because it does not deviate; the state is either -1 or 1

if we take your sequnece:

\[\{1,-\frac{1}{3},\frac{1}{5},-\frac{1}{7},\frac{1}{9},-\frac{1}{11},\frac{1}{13},-\frac{1}{15},...\}\]
we get a graph like this: