What is the sum from n=4 to infinity of 3-0.5^n ?

The sum from n=4 to infinity of 3-0.5^n is diverges

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sum from n=4 to infinity of 3-0.5^n

n = 4 \sum \infty 3 - 0. 5^{n}

diverges

Solution steps

n = 4 \sum \infty 3 - 0. 5^{n}

Apply Series Divergence Test:

diverges

= diverges

y^{^{'}} = \frac{x}{1 + 2 y}

\int \frac{20}{( x - 4 ) ( x ^{2} + 4 )} d x

sin (x), \frac{3 π}{2}

\frac{d}{d x} (100 x - x^{2})

n = 1 \sum \infty 3 n e^{- n^{2}}

What is the sum from n=4 to infinity of 3-0.5^n ?
The sum from n=4 to infinity of 3-0.5^n is diverges