What is the sum from n=1 to infinity}(16^{n/2 of)/(2^{2n)} ?

The sum from n=1 to infinity}(16^{n/2 of)/(2^{2n)} is diverges

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sum from n=1 to infinity}(16^{n/2 of)/(2^{2n)}

n = 1 \sum \infty \frac{1 6 ^{\frac{n}{2}}}{2 ^{2 n}}

diverges

Solution steps

n = 1 \sum \infty \frac{1 6 ^{\frac{n}{2}}}{2 ^{2 n}}

\frac{1 6 ^{\frac{n}{2}}}{2 ^{2 n}} = 1

= n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

\int \frac{1}{e ^{- y}} d y

d y = 2 x (y^{2} + 9) d x

\int 0.6 x d x

x \to \infty lim (\frac{4}{7})

f (x) = x^{2} e^{2 x}

What is the sum from n=1 to infinity}(16^{n/2 of)/(2^{2n)} ?
The sum from n=1 to infinity}(16^{n/2 of)/(2^{2n)} is diverges