Question

How to determine whether X_n converges or diverges?( by using limit theorem)
\(X_n= \dfrac{(-1)^n}{n+1}\)
Please, help

Answer 1

what limit theorem?

Answer 2

By definition of the limit, we can use \(lim |\dfrac{(-1)^n]{n+1}| = 0\)
but I don't know how to use limit theorem

Answer 3

If using limit theorem, we have to show Xn monotone and bounded

Answer 4

But X_n has 2 cases
\[X_n=\begin{cases}\dfrac{-1}{n+1}~~~if~~n~odd\\\dfrac{1}{n+1}~~~if~~n~even\end{cases}\]

Answer 5

Obviously, both them converge to 0, but how to put it in neat?

Answer 6

we have x1 < x2 (x1 is when n is odd <0, x2 when n even, >0)
we have x_k < x_(k+1) only if k is odd. if k even, we don't have it.
So .... I go nowhere.
Or can we use subsequences of X_n?

Answer 7

the first inequality you used both odd and even case

Answer 8

if n is odd we do have Xn<Xn+1

Answer 9

@SithsAndGiggles

Answer 10

May I ask what limit theorem are you referring to?

Answer 11

http://math.kennesaw.edu/~plaval/math4381/seqlimthm.pdf
maybe theorem 313 here

Answer 12

Let me scan

Answer 13

Well according to the definition on the page you scanned in if we could find M such that |xn|<M then xn is bounded.
Well we know for large n |xn|<1 so the sequence xn should be bounded.

Answer 14

Can I use theorem 3.1.3 ? by proving subsequences have the same limit then original one convergent?

Answer 15

the theorem3.1.3 from your link

Answer 16

ha, 3.1.5 , not 3.1.3

Answer 17

I go eat something, will be back later. Thanks for the help, though :)

Answer 18

I think you should be able to assume that x2n converges to 0. Then prove it.

Answer 19

\[|\frac{1}{n+1}-0|=|\frac{1}{n+1}|=\frac{1}{|n+1|} \\ =\frac{1}{n+1 } \text{ since } n \ge 1 \\ |\frac{1}{n+1}-0|=\frac{1}{n+1}<\frac{1}{n} \\ \frac{1}{n} < \epsilon => n>\frac{1}{\epsilon } \text{ So we have } |\frac{1}{n+1}-0|< \epsilon \text{ where } N=\frac{1}{\epsilon } . \]

Answer 20

http://tutorial.math.lamar.edu/Classes/CalcII/Sequences.aspx
the last theorem on pauls notes here is the one we are trying to apply

Answer 21

we already showed x2n->0