What is the sum from k=1 to infinity of 4/((2k-1)(2k+1)) ?

The sum from k=1 to infinity of 4/((2k-1)(2k+1)) is 2

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sum from k=1 to infinity of 4/((2k-1)(2k+1))

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

2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{2}

= 4 \cdot \frac{1}{2}

Simplify

4 \cdot \frac{1}{2} : 2

= 2

n = 0 \sum \infty ln (\frac{n}{3 n + 1})

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

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

n = 1 \sum \infty \frac{5}{n + 4}

n = 2 \sum \infty (- 0.95)^{n}

What is the sum from k=1 to infinity of 4/((2k-1)(2k+1)) ?
The sum from k=1 to infinity of 4/((2k-1)(2k+1)) is 2