What is the sum from n=1 to infinity of 6/(n^2) ?

The sum from n=1 to infinity of 6/(n^2) is converges

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sum from n=1 to infinity of 6/(n^2)

n = 1 \sum \infty \frac{6}{n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{n ^{2}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ^{2}}

Apply p-Series Test:

converges

= 6 converges

= converges

x = 2 \sum \infty \frac{1}{x} - \frac{1}{x - 1}

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{4}}

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

n = 2 \sum \infty \frac{1}{n ^{\frac{5}{3}}}

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

What is the sum from n=1 to infinity of 6/(n^2) ?
The sum from n=1 to infinity of 6/(n^2) is converges