Geometric Series

{sigma} 2*(-1/4)^(n-1)

Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2
my r= -1/4

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Mark44
Mentor
$$\sum_{n = 1}^{\infty} 2 (-1/4)^{n - 1}$$

Write a few terms of this series and see if you can adjust the starting index.

i have no idea what you mean by that

Hurkyl
Staff Emeritus
Gold Member
{sigma} 2*(-1/4)^(n-1)

Could i treat this as a geometric series? i know geometric is in the form of ar^n but the n is (n-1)

my A=2
my r= -1/4
You have
2(-1/4)n-1
and you want
A rn-1
?

Why not set them equal and try solving for A and r?

no... im trying to solve whether or not the series converges or diverges and then i have to determine the sum of the series... meaning where the series sums up to..

Hurkyl
Staff Emeritus
Gold Member
Oh, my mistake. I thought the method you had wanted to use was to rewrite the summand in a standard form, and were stuck on how to do that rewrite.

Char. Limit
Gold Member
Well, just change n-1 into n using exponent rules.

Mark44
Mentor
Write a few terms of the series and you should quickly see that it is a geometric series. A geometric series is one for which each term is a constant multiple of the previous term.