What is the sum from i=1 to infinity of 2(4/i)^3+9 ?

The sum from i=1 to infinity of 2(4/i)^3+9 is diverges

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sum from i=1 to infinity of 2(4/i)^3+9

i = 1 \sum \infty 2 (\frac{4}{i})^{3} + 9

diverges

Solution steps

i = 1 \sum \infty 2 (\frac{4}{i})^{3} + 9

Apply Series Divergence Test:

diverges

= diverges

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

n = 1 \sum \infty cos (\frac{9}{n})

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

n = 2 \sum \infty \frac{6}{n ln ( n )}

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

What is the sum from i=1 to infinity of 2(4/i)^3+9 ?
The sum from i=1 to infinity of 2(4/i)^3+9 is diverges