1: divergent, the term is greater than (3/2)^k, so every the sum is constantly increasing by an increasing amount

#3 & #7 are convergent because every term is less than 1/(n^2) and that series is convergent

if #8 is 1/(k^5) the same argument hiolds, but the picture is a little unclear

#5 is similar to #1 above

#6 is divergent because every term is larger than 1/k and that series diverges

#4 can be reordered so that every term is larger than 1/sqrt(n) which is larger than 1/n, so the series is divergent as well

sorry no comment on #2, don't have any reference material on hand - you have to wait for 'Azrad' or someone with similar knowledge/tools

This post was edited by brmv on Apr 29 2013 12:47am

Can anyone finish #2?

i would assume #2 diverges since it is an oscillating function. I'm not certain since your k-values are natural numbers (not continuous) and are taken at different points on the sine function, but that's my best educated guess.
I would seek a second opinion to be sure!

while it definitely is not straight forward convergent it could very well be conditionally convergent (which would be my educated guess)

Be extremely careful when you "reorder" terms in a serie.

But the conclusion is correct, #4 diverges.

You can simply check that general term is larger than 1 / k, that have a divergent serie.



#2 diverges simply because sin(k) doesn't tend to 0.

Having the general term tending to zero is a necessary (but not sufficient) condition for a serie to converge.

if you can reorder a series consisting of only positive terms so that it diverges it is safe to do so
one has to be careful when the series does not consist of all positive terms or one does a massive reorder to not forget anything, for #4 it is easy