# calculus

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I'm having trouble with a geometric series problem.

Determine if

the infinite summation of

(-3)^(n-1)/4^n

converges or diverges. If it converges, find the sum.

So the answer says that

sigma -3^(n-1)/4^n

= 1/4 * sigma (-3/4)^(n-1)

How did they factor out the 1/4? My algebra is very weak :(

• calculus -

the first part:

4^n can be written as
4(4)^(n-1) = (1/4) (4)^(n-1)

then (-3)^(n-1)/4^n
= (1/4) (-3)^(n-1) / 4^(n-1)
= (1/4) (-3/4)^(n-1)

no now:
∑ (1/4) (-3/4)^(n-1)

= (1/4)(-3/4)^0 + (1/4)(-3/4)^1 + (1/4)(-3/4)^2 + ...

which is an infinitite geometric series , which clearly converges.
with a = (1/4) and r = -3/4

sum∞ = a/(1-r) = (/4)/(1-(-3/4)
= (1/4)/(7/4)
= 1/7

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