What is the sum from n=1 to infinity of 5/(3^n+1) ?

The sum from n=1 to infinity of 5/(3^n+1) is converges

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sum from n=1 to infinity of 5/(3^n+1)

n = 1 \sum \infty \frac{5}{3 ^{n} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{5}{3 ^{n} + 1}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{3 ^{n} + 1}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 0 \sum \infty (2 n - 3) (4 x)^{n}

f (t) = 1.7 e^{0.76 t}

\frac{d}{d x} ((1 - x^{2}))

\int \frac{( x - 1 ) ^{3}}{x} d x

f (x) = 3 x^{2} + 1

What is the sum from n=1 to infinity of 5/(3^n+1) ?
The sum from n=1 to infinity of 5/(3^n+1) is converges