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

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

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

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

diverges

Solution steps

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

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}

Apply Series Divergence Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 0 \sum \infty \frac{2}{n + 2}

n = 0 \sum \infty \frac{1}{16 n ^{2} - 4}

n = 1 \sum \infty (0.7)^{n - 1}

n = 0 \sum \infty \frac{1}{n + 2} - \frac{1}{n}

n = 0 \sum \infty \frac{e ^{n}}{n ^{4}}

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