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

k = 2 \sum \infty (\frac{1}{4})^{2 k}

n = 1 \sum \infty \frac{9 n}{6 n + 1}

0 = 1 \sum \infty \frac{ln ( 0 )}{0}

n = 1 \sum \infty \frac{n ^{4}}{7 ^{n}}

n = 0 \sum \infty (\frac{5}{4})^{n - 1}

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