1. ## alternating series

i'm supposed to be testing if this series converges or diverges

$\displaystyle \sum_{n =1}^{\infty}(-1)^j\frac{\sqrt j}{j+5}$

i'm assuming i should be using the alternate series test here

so.. i'm taking the limit, and getting lim as n->infinity of (j^1/2)/j+5 is eventually equal to (1/2 j^-1/2 )/ 6. i think that this limit equals zero, is that correct?

also, i'm wondering how i can easily prove that bn+1 < or equal to bn

thanks!

2. Originally Posted by buttonbear
i'm supposed to be testing if this series converges or diverges

$\displaystyle \sum_{n =1}^{\infty}(-1)^j\frac{\sqrt j}{j+5}$

i'm assuming i should be using the alternate series test here

so.. i'm taking the limit, and getting lim as n->infinity of (j^1/2)/j+5 is eventually equal to (1/2 j^-1/2 )/ 6. i think that this limit equals zero, is that correct?

also, i'm wondering how i can easily prove that bn+1 < or equal to bn

thanks!
There are j's where there should be n's (or vice versa) and I can't follow your argument.

All you have to do is prove $\displaystyle \lim_{n \rightarrow +\infty} \frac{\sqrt{n}}{n + 5} = 0$.

Note that $\displaystyle \frac{\sqrt{n}}{n + 5} = \frac{1}{\sqrt{n} + \frac{5}{\sqrt{n}}}$ and the result is obvious.

3. Originally Posted by mr fantastic
There are j's where there should be n's (or vice versa) and I can't follow your argument.

All you have to do is prove $\displaystyle \lim_{n \rightarrow +\infty} \frac{\sqrt{n}}{n + 5} = 0$.

Note that $\displaystyle \frac{\sqrt{n}}{n + 5} = \frac{1}{\sqrt{n} + \frac{5}{\sqrt{n}}}$ and the result is obvious.
Not quite, you also have to show that eventually $\displaystyle \frac{\sqrt{n}}{n + 5}$ is decreasing.

CB

4. yeah..they're all supposed to be j's, sorry about that

and, you do have to do more than just show the limit goes to zero

you have to show it is decreasing, that was what i was asking about
a clear way to demonstrate that bj+1 < or equal to bj

5. Originally Posted by buttonbear
you do have to do more than just show the limit goes to zero you have to show it is decreasing, that was what i was asking about
Look at the derivative of $\displaystyle \frac{\sqrt{x}}{x+5}$ you will see that it is negative (decreasing) if $\displaystyle 5<x$.