What is the sum from n=0 to infinity of (n^2)/(5+2n) ?

The sum from n=0 to infinity of (n^2)/(5+2n) is diverges

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sum from n=0 to infinity of (n^2)/(5+2n)

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

diverges

Solution steps

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

Apply Series Divergence Test:

diverges

= diverges

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

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

i = 1 \sum \infty 5 \cdot 0. 2^{i - 1}

n = 1 \sum \infty \frac{3}{5 ^{n + 2}}

n = 0 \sum \infty ln (1 - n^{2})

What is the sum from n=0 to infinity of (n^2)/(5+2n) ?
The sum from n=0 to infinity of (n^2)/(5+2n) is diverges