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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 1 \sum \infty \frac{2}{3 n}

n = 1 \sum \infty \frac{2 n}{7 n - 4}

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

n = 1 \sum \infty \frac{n ^{2}}{n ^{4}}

n = 5 \sum \infty \frac{3}{n ^{2} + 5 n + 4}

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