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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