What is the sum from n=1 to infinity}(\sqrt{2 of)^n ?

The sum from n=1 to infinity}(\sqrt{2 of)^n is diverges

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sum from n=1 to infinity}(\sqrt{2 of)^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

n = 0 \sum \infty 8 - 5 n^{4}

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

n = 0 \sum \infty \frac{n ^{2}}{7 ^{n}}

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

n = 1 \sum \infty \frac{8}{e ^{n} - 1}

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