What is the sum from n=1 to infinity of 1/(2^n-5) ?

The sum from n=1 to infinity of 1/(2^n-5) is converges

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sum from n=1 to infinity of 1/(2^n-5)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d y}{d t} + 0.7 t y = 7 t

y = 5 x^{2} + 2 x y - 3 y^{2} + 3, (2, 1)

\frac{\partial}{\partial x} (1 - 9 x^{2} - 6 y^{2})

f (x) = (x + 2) \cdot - x

\frac{d y}{d x} = \frac{2 y}{x} - x^{2} y^{2}

What is the sum from n=1 to infinity of 1/(2^n-5) ?
The sum from n=1 to infinity of 1/(2^n-5) is converges