What is the sum from n=1 to infinity of 4/(3^n+1) ?

The sum from n=1 to infinity of 4/(3^n+1) is converges

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sum from n=1 to infinity of 4/(3^n+1)

n = 1 \sum \infty \frac{4}{3 ^{n} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{4}{3 ^{n} + 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{3 ^{n} + 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

y = 3 x^{2} + 2^{2} x^{2} + 1, (- 1, 0)

\int \frac{- 1}{x + 4} d x

f (x) = 6 + ln (x), at x = 1

\frac{\partial}{\partial x} (e^{3 x} (x^{2} + y^{2}))

e^{- 5 x} + e^{2 x}

What is the sum from n=1 to infinity of 4/(3^n+1) ?
The sum from n=1 to infinity of 4/(3^n+1) is converges