Solving Geometric Series: 2*(-1/4)^(n-1)

[mattmannmf]

{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

 

[Mark44]

Your series appears to be
\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.

 

[mattmannmf]

i have no idea what you mean by that

 

[Hurkyl]

{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 r^n-1​

?

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

 

[mattmannmf]

no... I am 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]

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]

Well, just change n-1 into n using exponent rules.

 

[Mark44]

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.