What is the sum from n=1 to infinity of 4/n ?

The sum from n=1 to infinity of 4/n is diverges

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sum from n=1 to infinity of 4/n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Cauchy's Convergence Condition:

diverges

= 4 diverges

= diverges

\int \frac{3}{2 x - 1} d x

y^{^{'}} - 1 0^{- 3} y = 220 cos (50 x)

d y - k y d x = 0

x \to 1 lim (\frac{2 x + 1}{x - 1})

n = 7 \sum \infty (\frac{6}{7})^{- n}

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