What is the sum from n=1 to infinity of 3n^{-2} ?

The sum from n=1 to infinity of 3n^{-2} is converges

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sum from n=1 to infinity of 3n^{-2}

n = 1 \sum \infty 3 n^{- 2}

converges

Solution steps

n = 1 \sum \infty 3 n^{- 2}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty n^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply p-Series Test:

converges

= 3 converges

= converges

n = 0 \sum \infty \frac{n ^{4}}{n + 2}

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

n = 1 \sum \infty \frac{n ^{5}}{n ^{7}}

n = 1 \sum \infty \frac{1}{ln ( 3 ^{n} )}

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

What is the sum from n=1 to infinity of 3n^{-2} ?
The sum from n=1 to infinity of 3n^{-2} is converges