What is the sum from n=1 to infinity of 2/(7^n+1) ?

The sum from n=1 to infinity of 2/(7^n+1) is converges

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sum from n=1 to infinity of 2/(7^n+1)

n = 1 \sum \infty \frac{2}{7 ^{n} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{2}{7 ^{n} + 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{7 ^{n} + 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 0 \sum \infty 10 \cdot 9.14

n = 1 \sum \infty \frac{n ^{2}}{5 n ^{2}}

n = 0 \sum \infty (n!)^{2} (0.5)

k = 0 \sum \infty (- \frac{2}{5})^{k}

n = 1 \sum \infty 8 n + 1

What is the sum from n=1 to infinity of 2/(7^n+1) ?
The sum from n=1 to infinity of 2/(7^n+1) is converges