What is the sum from n=1 to infinity of n/(4n+1) ?

The sum from n=1 to infinity of n/(4n+1) is diverges

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sum from n=1 to infinity of n/(4n+1)

n = 1 \sum \infty \frac{n}{4 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{4 n + 1}

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial y} (x^{3} - y^{3} + 6 x y)

\frac{d}{d x} (x y^{2} - x^{3} y) = 0

(- 21.2, 22.8), (- 20.2, 7.4)

n = 1 \sum \infty \frac{5}{π ^{n}}

\int \frac{( x + 4 )}{x ^{2} + 11 x + 18} d x

What is the sum from n=1 to infinity of n/(4n+1) ?
The sum from n=1 to infinity of n/(4n+1) is diverges