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

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

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

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

converges

Solution steps

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

Apply p-Series Test:

converges

= \frac{1}{2} converges

= converges

x^{2} e^{x} - x^{3} - x^{2}

\frac{d}{d t} (2 e^{3 t})

\frac{d}{d x} (\frac{sin ( x )}{7 + cos ( x )})

\frac{d}{d x} (5 + 6 x^{\frac{3}{2}})

x^{^{''}} + 4 x = 3

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