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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{3 ^{n}}{1 0 ^{n}}

n = 1 \sum \infty e^{- n} + e^{n}

n = 1 \sum \infty \frac{1}{n ! 4 ^{n}}

n = 0 \sum \infty \frac{1}{3 - n}

n = 1 \sum \infty \frac{3}{2 n}

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