What is the sum from n=2 to infinity of 4n^{-2} ?

The sum from n=2 to infinity of 4n^{-2} is converges

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sum from n=2 to infinity of 4n^{-2}

n = 2 \sum \infty 4 n^{- 2}

converges

Solution steps

n = 2 \sum \infty 4 n^{- 2}

Apply the constant multiplication rule:

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

= 4 \cdot n = 2 \sum \infty n^{- 2}

Apply exponent rule:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{n ^{2}}

Apply p-Series Test:

converges

= 4 converges

= converges

n = 0 \sum \infty (\frac{1}{13})^{n}

n = 6 \sum \infty \frac{1}{( n - 5 ) ^{2}}

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

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

n = 1 \sum \infty \frac{π}{n ^{2}}

What is the sum from n=2 to infinity of 4n^{-2} ?
The sum from n=2 to infinity of 4n^{-2} is converges