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

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

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

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

converges

Solution steps

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

Apply Alternating Series Test:

converges

= converges

\frac{\partial}{\partial x} (\frac{1}{1 - x})

x d x - y^{2} d y = 0

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

x \to 5 - lim (\frac{3}{5 - x})

\frac{\partial}{\partial y} (y + y ln (x y))

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