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

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

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sum from n=0 to infinity of n/(1-4n)

n = 0 \sum \infty \frac{n}{1 - 4 n}

diverges

Solution steps

n = 0 \sum \infty \frac{n}{1 - 4 n}

Factor

1 - 4 n : - (- 1 + 4 n)

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 0 \sum \infty \frac{n}{- 1 + 4 n}

Apply Series Divergence Test:

diverges

= - diverges

= diverges

n = 16 \sum \infty \frac{1}{ln ( n )}

n = 1 \sum \infty \frac{1}{9 n}

k = 2 \sum \infty (\frac{3}{5})^{3 k}

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

k = 1 \sum \infty \frac{1}{2 k}

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