What is the sum from k=1 to infinity}((-1)^k*(1)^{(3*k) of)/k ?

The sum from k=1 to infinity}((-1)^k*(1)^{(3*k) of)/k is converges

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sum from k=1 to infinity}((-1)^k(1)^{(3k) of)/k

k = 1 \sum \infty \frac{( - 1 ) ^{k} \cdot ( 1 ) ^{(3 \cdot k)}}{k}

converges

Solution steps

k = 1 \sum \infty \frac{( - 1 ) ^{k} \cdot 1 ^{3 k}}{k}

Simplify

= k = 1 \sum \infty \frac{( - 1 ) ^{k}}{k}

Apply Alternating Series Test:

converges

= converges

\int x y (x + y) d y

n = 0 \sum \infty \frac{2 ^{n}}{n !}

x (x - 2) y^{^{'}} + 2 y = 0

\int (x + 2) x - 2 d x

\int \frac{1}{( x ^{4} - 1 )} d x

What is the sum from k=1 to infinity}((-1)^k*(1)^{(3*k) of)/k ?
The sum from k=1 to infinity}((-1)^k*(1)^{(3*k) of)/k is converges