1. ## Help about ratio test for series..

I have a series and i have a problem doing the ratio test....

Σ k=1 to infinity, (-1)^k (1/2 + 1/k)^k

i face my problem with the 1/k :/

edit: (i have to actually examine the absolute convergence and i think i have to first examine the ratio..)

2. ## Re: Help about ratio test for series..

Originally Posted by nappysnake
I have a series and i have a problem doing the ratio test....

Σ k=1 to infinity, (-1)^k (1/2 + 1/k)^k

i face my problem with the 1/k :/
Use the alternating series test (1/2 + 1/k)^k is monotone decreasing after a a certain value of n ....)

3. ## Re: Help about ratio test for series..

The ratio test will work too. By the ratio test, the series will be convergent if \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*}

\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n \to \infty}\left|\frac{(-1)^{n+1}\left(\frac{1}{2} + \frac{1}{n+1}\right)^{n+1}}{(-1)^n\left(\frac{1}{2} + \frac{1}{n}\right)^n}\right| \\ &= \lim_{n \to \infty}\frac{\left(\frac{1}{2} + \frac{1}{n+1}\right)^{n+1}}{\left(\frac{1}{2} + \frac{1}{n}\right)^n} \\ &= \frac{1}{2} \textrm{ according to Wolfram Alpha.} \end{align*}

So the series converges.

4. ## Re: Help about ratio test for series..

i have one further question. how would you calculate absolute convergence? wolfram alpha says that the ratio test is conclusive and that the series converges, but i have no idea how he comes to the result..all i get is a bunch of terms which i can't simplify..help?

5. ## Re: Help about ratio test for series..

Originally Posted by nappysnake
I have a series and i have a problem doing the ratio test....

Σ k=1 to infinity, (-1)^k (1/2 + 1/k)^k

i face my problem with the 1/k :/

edit: (i have to actually examine the absolute convergence and i think i have to first examine the ratio..)
It is difficult to undestand why it is requested the ratio test instead of the root test, that extablishes that a series $\displaystyle \sum_{n=0}^{\infty} a_{n}$ converges if...

$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n] {|a_{n}|} <1$ (1)

In that case is...

$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n] {|a_{n}|} = \lim_{n \rightarrow \infty} (\frac{1}{2} + \frac{1}{n})= \frac{1}{2}$ (2)

Marry Christmas from Serbia

$\displaystyle \chi$ $\displaystyle \sigma$