What is the sum from n=0 to infinity of 2+4+6+8 ?

The sum from n=0 to infinity of 2+4+6+8 is diverges

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sum from n=0 to infinity of 2+4+6+8

n = 0 \sum \infty 2 + 4 + 6 + 8

diverges

Solution steps

n = 0 \sum \infty 2 + 4 + 6 + 8

Every infinite sum of a non-zero constant diverges

= diverges

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\frac{d}{d x} ((x^{2} + 7)^{3})

x \to - 10 lim (3 x - 8)

\frac{\partial}{\partial r} (r sin (x))

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What is the sum from n=0 to infinity of 2+4+6+8 ?
The sum from n=0 to infinity of 2+4+6+8 is diverges