find sum of series?

Question

........summation sign.....(-3)^n-1 / (5) ^n-1

Answer 1

.....(-3)^n-1 / (5) ^n-1
I assume you mean summation from n = 1 to n = ∞ and the series is (-3)^(n-1) / (5) ^(n-1)

= Σ (n =1 to n=∞) [ (-3)^(n-1) / (5) ^(n-1) ]
= Σ (n =1 to n=∞) ( -3/5)^(n -1)
= 1 - 3/5 + (3/5)^2 - (3/5)^3 + .... to ∞
= 1 / (1 + 3/5)
= 5/8.

Answer 2

The 1st few terms are 1, -3/5, 9/25, -27/125 + 81/625 -243/3125. This is an alternating geometric series with r = 3/5
The first 6 terms add to .59584 which has an error < .046656.
The error in taking the sum of the first n terms as an approximation to the sum is numerically less than the (n+1)st term.

Answer 3

I assume this has n from 1 to inf. Is this not a geometric series? For, the numerator and denominator can be combined, so that the general term may be written

(-3/5)^(n-1).

According to the geometric series formula, this series converges, and its sum is

1/(1 + 3/5) = 5/8.