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

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

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sum from n=1 to infinity of 6/n

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

diverges

Solution steps

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

\frac{d}{d x} (2 + 6 x)

\frac{d}{d y} (y^{y})

\frac{d}{d x} (e^{3 x^{2}} + ln (5 x - 4))

\frac{d}{d x} (x - sin^{2} (x) - 4)

\frac{4 sin ( x )}{6 - 7 sin ( x )}

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