Ask your own question, for FREE!
Mathematics
OpenStudy (loser66):

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

OpenStudy (xapproachesinfinity):

what limit theorem?

OpenStudy (loser66):

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

OpenStudy (loser66):

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

OpenStudy (loser66):

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

OpenStudy (loser66):

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

OpenStudy (loser66):

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?

OpenStudy (xapproachesinfinity):

the first inequality you used both odd and even case

OpenStudy (xapproachesinfinity):

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

OpenStudy (xapproachesinfinity):

@SithsAndGiggles

myininaya (myininaya):

May I ask what limit theorem are you referring to?

myininaya (myininaya):

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

OpenStudy (loser66):

Let me scan

myininaya (myininaya):

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.

OpenStudy (loser66):

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

OpenStudy (loser66):

the theorem3.1.3 from your link

OpenStudy (loser66):

ha, 3.1.5 , not 3.1.3

OpenStudy (loser66):

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

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

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

myininaya (myininaya):

we already showed x2n->0

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!