What is the integral from 1 to infinity of 1/(2^n) ?

The integral from 1 to infinity of 1/(2^n) is diverges

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integral from 1 to infinity of 1/(2^n)

\int_{1}^{\infty} \frac{1}{2 ^{n}}

diverges

Solution steps

\int_{1}^{\infty} \frac{1}{2 ^{n}} d x

Compute the indefinite integral:

\int \frac{1}{2 ^{n}} d x = \frac{x}{2 ^{n}} + C

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{2 ^{n}} d x =

diverges

= diverges

\int_{0}^{1} - 2 x^{2} d x

\int_{0}^{1} x (x^{2} + 2) d x

\int_{- k}^{k} ∣ k x ∣ d x

\int_{0}^{1} e^{x} - e^{2 x - 1} d x

\int_{0}^{\frac{3}{2}} \frac{x - 3}{2 x - 3} d x

What is the integral from 1 to infinity of 1/(2^n) ?
The integral from 1 to infinity of 1/(2^n) is diverges