What is the sum from n=1 to infinity of 2^n ?

The sum from n=1 to infinity of 2^n is diverges

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sum from n=1 to infinity of 2^n

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

diverges

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty 2^{n} - 2^{0}

n = 0 \sum \infty 2^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{\frac{π}{4}} 2 sec (t) d t

n = 1 \sum \infty (- 1)^{n} 9^{n} x^{n}

\frac{d}{d x} (1 + x - 1)

y = x^{2} - x^{4}, (1, 0)

\int_{1}^{6} \frac{7}{x ^{2}} d x

What is the sum from n=1 to infinity of 2^n ?
The sum from n=1 to infinity of 2^n is diverges