What is the sum from n=1 to infinity of 6/(4n^2-1) ?

The sum from n=1 to infinity of 6/(4n^2-1) is 3

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sum from n=1 to infinity of 6/(4n^2-1)

n = 1 \sum \infty \frac{6}{4 n ^{2} - 1}

3

Solution steps

n = 1 \sum \infty \frac{6}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 6 \cdot \frac{1}{2}

Simplify

6 \cdot \frac{1}{2} : 3

= 3

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

k = 1 \sum \infty k e^{- 3 k}

n = 1 \sum \infty 2 sin (\frac{2}{n})

n = 1 \sum \infty \frac{1}{( n + 3 ) !}

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

What is the sum from n=1 to infinity of 6/(4n^2-1) ?
The sum from n=1 to infinity of 6/(4n^2-1) is 3