What is the sum from n=0 to infinity of n/((1+4n)^2) ?

The sum from n=0 to infinity of n/((1+4n)^2) is diverges

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sum from n=0 to infinity of n/((1+4n)^2)

n = 0 \sum \infty \frac{n}{( 1 + 4 n ) ^{2}}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{( 1 + 4 n ) ^{2}}

Apply Series Comparison Test:

diverges

= diverges

n = 1 \sum \infty cos (8 n)

n = 0 \sum \infty n^{^{'}}

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

n = 1 \sum \infty 2 (\frac{3}{4})^{n - 1}

n = 1 \sum \infty (sin (100))^{n}

What is the sum from n=0 to infinity of n/((1+4n)^2) ?
The sum from n=0 to infinity of n/((1+4n)^2) is diverges