What is the sum from n=0 to infinity of (n+8)^2 ?

The sum from n=0 to infinity of (n+8)^2 is diverges

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sum from n=0 to infinity of (n+8)^2

n = 0 \sum \infty (n + 8)^{2}

diverges

Solution steps

n = 0 \sum \infty (n + 8)^{2}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{- 3 n}{2 n + 1}

n = 2 \sum \infty \frac{1}{n ^{5}}

n = 1 \sum \infty \frac{1}{( n ^{2} )}

n = 0 \sum \infty \frac{3 n + 1}{2 n}

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

What is the sum from n=0 to infinity of (n+8)^2 ?
The sum from n=0 to infinity of (n+8)^2 is diverges