What is the sum from n=1 to infinity of 6 1/n ?

The sum from n=1 to infinity of 6 1/n is diverges

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sum from n=1 to infinity of 6 1/n

n = 1 \sum \infty 6 \frac{1}{n}

diverges

Solution steps

n = 1 \sum \infty 6 \cdot \frac{1}{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 6 diverges

= diverges

f (x) = x^{2} + 3 x + 4 π, at x = - 3

\frac{d}{d x} (arccos (\frac{x}{b}))

\frac{d ^{4}}{d x ^{4}} (\frac{8 x}{1 - x})

f (x) = 5^{5 x + 1} + 3^{1 - 3 x}

y^{^{'}} + 5 y = 7

What is the sum from n=1 to infinity of 6 1/n ?
The sum from n=1 to infinity of 6 1/n is diverges