# Simple Question

• Mar 19th 2009, 06:38 AM
bearej50
Simple Question
sn = (((-1)^n)*n)/(2n - 1)

How would I determine the convergence or divergence of the sequence (sn) and find the limit if it exists?
• Mar 19th 2009, 07:39 AM
CaptainBlack
Quote:

Originally Posted by bearej50
sn = (((-1)^n)*n)/(2n - 1)

How would I determine the convergence or divergence of the sequence (sn) and find the limit if it exists?

As this sequence has alternating signs it can only converge to $\displaystyle 0$ and so $\displaystyle |s_n|$ must also converge to $\displaystyle 0$. For $\displaystyle n>0$:

$\displaystyle |s_n|=n/(2n-1)=1/(2-1/n) \to 1/2$ as $\displaystyle n \to \infty$

so the original sequence does not converge.

CB
• Mar 19th 2009, 07:53 AM
Showcase_22
Quote:

$\displaystyle S_n=\frac{n(-1)^n}{2n-1}$
I would see if this has a limit by the ratio test:

$\displaystyle \frac{S_{n+1}}{S_n}=\frac{(n+1)(-1)^{n+1}}{2(n+1)-1}. \frac{2n-1}{n(-1)^n}=\frac{-(n+1)(2n-1)}{n(2n+1)}=-(1+\frac{1}{n})(\frac{2n-1}{2n+1})$

$\displaystyle =-(1+\frac{1}{n})(\frac{2n+1-2}{2n+1})=-(1+\frac{1}{n})(1-\frac{2}{2n+1})=-(1+\frac{1}{n}-\frac{2}{2n+1}-\frac{2}{n(2n+1)})$$\displaystyle \rightarrow -1$

but for the ratio test (I should have probably put this in before =S) we need to take the modulus.

$\displaystyle |\frac{S_{n+1}}{S_n}| \rightarrow 1$

Unfortunately, if the ratio test produces 1 the test doesn't hold. =S