What is the sum from n=1 to infinity of (2n)/(n^2) ?

The sum from n=1 to infinity of (2n)/(n^2) is diverges

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sum from n=1 to infinity of (2n)/(n^2)

n = 1 \sum \infty \frac{2 n}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{n ^{2}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{n ^{2}}

Cancel the common factor:

n

= 2 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 2 diverges

= diverges

f (x) = 7 arctan (7 x)

y = - 8 x ln (6 x + 5)

\int_{0}^{1} (x + 3) (x - 4) d x

y = (x^{3} - 16 x)^{12}, (4, 0)

\frac{d}{d y} (arctan (x + 2 y))

What is the sum from n=1 to infinity of (2n)/(n^2) ?
The sum from n=1 to infinity of (2n)/(n^2) is diverges