need help with these series

Question

anonymous · Answer

$$\sum_{n=2}^{infity} 3^(-n)8^(n+1)$$

anonymous · Answer

hard to show i guess but both are to that power,-n, n+1

rogue · Answer

$$\sum_{n=2}^{\infty} 3^{-n} 8 ^{n+1}$$Is that your series?

rogue · Answer

If so, it is divergent.

anonymous · Answer

Yes
how do I show that?

rogue · Answer

$$\sum_{n=2}^{\infty} 3^{-n} 8 ^{n+1} = \sum_{n=2}^{\infty} 8 \times \frac {8^n}{3^n} = 8 \sum_{n=2}^{\infty}( \frac {8}{3})^n$$You don't really have to show anything for this one. Just say that this is a geometric series with a common ratio of 8/3, which greater than 1, and so the series diverges.

anonymous · Answer

ahhh i would have never seen it that way!  Thanks, really helps

anonymous · Answer

What about this one quick...

rogue · Answer

Your welcome =)
I hated sequences and series at first, but now I love them :)

anonymous · Answer

$$\sum_{n=1}^{inifinity}n/(n+10)$$

anonymous · Answer

Im struggling so far.. I never know how to approach. Like that one wouldnt I take the limit?

rogue · Answer

Yeah, show that the limit doesn't = 0.

anonymous · Answer

Oh so because it is equal to 1/10 we know it diverges?

anonymous · Answer

Or that was stupid...

rogue · Answer

The limit is 1 =P

anonymous · Answer

I mean it would equal 1/(10/n), 10/n goes to 0, so limit =1/1=1

anonymous · Answer

Yeah I just forgot to divide 10 by n also. Thanks