What is the sum from n=1 to infinity of n/(1-7n) ?

The sum from n=1 to infinity of n/(1-7n) is diverges

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sum from n=1 to infinity of n/(1-7n)

n = 1 \sum \infty \frac{n}{1 - 7 n}

diverges

Solution steps

n = 1 \sum \infty \frac{n}{1 - 7 n}

Factor

1 - 7 n : - (- 1 + 7 n)

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

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{n}{- 1 + 7 n}

Apply Series Divergence Test:

diverges

= - diverges

= diverges

\frac{d}{d x} (\frac{- 6 x}{2 x - 1})

\frac{4 x ^{5} + 7}{x ^{3}}

\frac{d}{d x} (ln (\frac{x ^{2} + 1}{x + 4}))

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 2, y (1) = 3

x \to 7 + lim (ln (x - 7))

What is the sum from n=1 to infinity of n/(1-7n) ?
The sum from n=1 to infinity of n/(1-7n) is diverges