What is the sum from n=0 to infinity of (2n)/(n^2) ?

The sum from n=0 to infinity of (2n)/(n^2) is diverges

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sum from n=0 to infinity of (2n)/(n^2)

n = 0 \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

\frac{\partial}{\partial y} ((x y^{2} + x^{3})^{4})

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

f (x) = 5 x^{2} - 5 x, at x = 3

\int \frac{arctan ( \frac{x}{2} )}{4 + x ^{2}} d x

\frac{d}{d x} (x^{3} - x + 1)

What is the sum from n=0 to infinity of (2n)/(n^2) ?
The sum from n=0 to infinity of (2n)/(n^2) is diverges