Geometric and alternating series.?

Question

How can I tell apart a geometric series from an alternating? Example: ∑ ((-6)^n-1)/(5^n-1), where n =1 and goes to infinity.

∑((-1)^n-1)/(n^6), where n=1 and goes to infinity.

Which is which? Should I just try to identify a and r, and if I get them, then is a geomtric? In other words, if I find the ratio, that proves the series to be geometric?

Answer 1

A geometric series is one for which An+1/ An is constant

This is the case for the first series you put down, so it is geometric.

If An+1/ An is a function of n, you do not a a Geometric series.

The second series is a p-series.

Answer 2

The first one is S (-6/5)^n *( 1-1/(-6)^n)/1-1/5^n)
The second factor==>1 and the first factor does not ==> to 0 so the product does not ==>0 and the series is divergent
2) Taken in abs val a_n = 1/n^6 which by the integral tes is convergent
S a_n < Int ( 1,infinity )1/x^6 dx which is convergent
so the alternate series is absolutely convergent
Observation if (n-1) is the whole exponent the first is a geometric series with r=(-6/5) <-1 so it is divergent

Answer 3

(-6)^(n-1)/(5^(n-1)

(-6)^0/5^0, (-6)^1/5^1, (-6)^2/5^2

1/1, -6/5, 36/25,

This is geometric series with

a = 1 and r = -6/5

The sum of the series upto infinite is given by

S(inf) = a/(1-r) = 1/(1+6/5) = 1/(11/5) = 5/11