What is the sum from n=1 to infinity of 3/(2^n+1) ?

The sum from n=1 to infinity of 3/(2^n+1) is converges

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sum from n=1 to infinity of 3/(2^n+1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{2 ^{n} + 1}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int \frac{ln ( 4 x )}{x ^{3}} d x

f (x) = \frac{( x ^{2} + 9 )}{( x ^{2} - 9 )}

\int_{b}^{2 b} x^{3} d x

6 + x - x^{2}, 4

\int \frac{( x ^{2} )}{4 - x ^{2}} d x

What is the sum from n=1 to infinity of 3/(2^n+1) ?
The sum from n=1 to infinity of 3/(2^n+1) is converges