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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Every infinite sum of a non-zero constant diverges

= \frac{1}{2} diverges

= diverges

y = 4 x^{2} sin (x) tan (x)

y^{^{'}} = \frac{1}{t y + 2 t + y + 2}

n \to a lim (\frac{5 n}{10 n + 4})

\frac{d}{d x} (5 - cos (x))

\frac{\partial}{\partial y} (\frac{x ^{2} y - y}{x ^{2} - 2 y})

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