hey mate,

there are numerous ways to test converge of sums of infinite series, however the most common of which is to use the ratio test,

Ratio Test :

Given an infinite sum of the form

S = sum (a(n), n = 0..inf)

then if L < 1, the series converges, L > 1 the series diverges and finally if L = 1 the test fails and nothing can be stated (by the Ratio test) about its convergence,

where

L = lim (n--> inf) |a(n+1)/a(n)|

As an example I will do one of your questions where

a(n) = (-1)^n * (1/n)*(1/ln(n+3))

then a(n+1) = (-1)^(n+1) * (1/(n+1))*(1/ln(n + 4))

thus,

a(n+1)/a(n) = (-1)^(n+1) * (1/(n+1))*(1/ln(n + 4)) / ((-1)^n * (1/n)*(1/ln(n+3)))

= (-1)*(n/(n+1))*(ln(n+3)/ln(n+5))

thus,

|a(n+1)/a(n)| = (n/(n+1))*(ln(n+3)/ln(n+5))

we now need to evaluate the limit,

L = lim (n --> inf) (n/(n+1))*(ln(n+3)/ln(n+5))

note it is easily observed that lim (n--> inf) (n/(n+1)) = 1 then L becomes

L = lim (n --> inf) ln(n+3)/ln(n+5)

= infinity/infinity

Hence employ L'hopitals Rule

L = lim (n--> inf) d/dn (ln(n+3))/ d/dn (ln(n+5))

L = lim (n--> inf) (1/(n+3))/(1/(n+5))

L = lim (n--> inf) (n+5)/(n+3) = 5/3 > 1

Thus the series diverges.

Hope this was informative and I didnt go overboard with the description.

Regards,

David

ps - forgot to mention that for something to be absolutely convergent then the sum( |a(n)| , n = 0..inf ) must converge, in order to determine this, simply take the absolute value of the integrand of summation a(n) and apply the Ratio test as prescribed.