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

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

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sum from n=1 to infinity of n/(9n^2+1)

n = 1 \sum \infty \frac{n}{9 n ^{2} + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{9 n ^{2} + 1}

Apply Series Integral Test:

diverges

= diverges

\int \frac{x ^{2} - 100}{x} d x

\frac{d y}{d x} = (y sin (x))

\int sin (12 x) cos (7 x) d x

2 t + 1

x^{2} + 6, 7 x^{2}

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