What is the sum from n=0 to infinity of 3/((1+i)^n) ?

The sum from n=0 to infinity of 3/((1+i)^n) is converges

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sum from n=0 to infinity of 3/((1+i)^n)

n = 0 \sum \infty \frac{3}{( 1 + i ) ^{n}}

converges

Solution steps

n = 0 \sum \infty \frac{3}{( 1 + i ) ^{n}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 0 \sum \infty \frac{1}{( 1 + i ) ^{n}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\frac{d y}{d x}, e^{y} = x

\frac{d}{d u} (u sin (v))

\frac{d}{d x} (3 + x^{2})

y^{^{''}} + 5 y = - 180 x^{2} e^{5} x

f (x) = x - ln (x)

What is the sum from n=0 to infinity of 3/((1+i)^n) ?
The sum from n=0 to infinity of 3/((1+i)^n) is converges