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?

Printable View

- Mar 19th 2009, 06:38 AMbearej50Simple Question
*s**n*= (((-1)^*n)***n*)/(2*n*- 1)

How would I determine the convergence or divergence of the sequence (*s**n*) and find the limit if it exists? - Mar 19th 2009, 07:39 AMCaptainBlack
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 AMShowcase_22Quote:

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

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

$\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