What is the sum from n=0 to infinity of 4/((1-4n)^2) ?

The sum from n=0 to infinity of 4/((1-4n)^2) is converges

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sum from n=0 to infinity of 4/((1-4n)^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{( 1 - 4 n ) ^{2}}

Apply Series Comparison Test:

converges

= 4 converges

= converges

n = 1 \sum \infty n^{5} e^{n^{3}}

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

n = 1 \sum \infty \frac{- 1}{8 ^{n}}

n = 0 \sum \infty e^{- n^{8}}

n = 2 \sum \infty \frac{4}{n ^{3}}

What is the sum from n=0 to infinity of 4/((1-4n)^2) ?
The sum from n=0 to infinity of 4/((1-4n)^2) is converges