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

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

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sum from k=1 to infinity of 2^k

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

diverges

Solution steps

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

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 0 \sum \infty \frac{4}{8 ^{n}}

n = 1 \sum \infty 3^{n - 1}

n = 0 \sum \infty \frac{2 ^{n}}{2 n - 1}

n = 1 \sum \infty 1 - \frac{1}{n + 1}

n = 1 \sum \infty \frac{1}{( 4 n + 3 ) ^{4}}

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