What is the sum from k=0 to infinity of 1/(900+3k) ?

The sum from k=0 to infinity of 1/(900+3k) is diverges

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sum from k=0 to infinity of 1/(900+3k)

k = 0 \sum \infty \frac{1}{900 + 3 k}

diverges

Solution steps

k = 0 \sum \infty \frac{1}{900 + 3 k}

Apply Series Integral Test:

diverges

= diverges

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

n = 1 \sum \infty 3 (- \frac{1}{7})^{3 n}

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

y^{^{'}} = 3 x y + 5 y

sin (3 x) d x + 2 y cos^{3} (3 x) d y = 0

What is the sum from k=0 to infinity of 1/(900+3k) ?
The sum from k=0 to infinity of 1/(900+3k) is diverges