What is the sum from n=1 to infinity of (n^2)/(5n^2) ?

The sum from n=1 to infinity of (n^2)/(5n^2) is diverges

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sum from n=1 to infinity of (n^2)/(5n^2)

n = 1 \sum \infty \frac{n ^{2}}{5 n ^{2}}

diverges

Solution steps

n = 1 \sum \infty \frac{n ^{2}}{5 n ^{2}}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{n ^{2}}{n ^{2}}

Simplify

= \frac{1}{5} \cdot n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= \frac{1}{5} diverges

= diverges

n = 0 \sum \infty (n!)^{2} (0.5)

k = 0 \sum \infty (- \frac{2}{5})^{k}

n = 1 \sum \infty 8 n + 1

n = 0 \sum \infty sin (e^{- n})

n = 1 \sum \infty 2 (- \frac{1}{7})^{2 n}

What is the sum from n=1 to infinity of (n^2)/(5n^2) ?
The sum from n=1 to infinity of (n^2)/(5n^2) is diverges