What is the sum from n=1 to infinity of 8/(4n^2-1) ?

The sum from n=1 to infinity of 8/(4n^2-1) is 4

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sum from n=1 to infinity of 8/(4n^2-1)

n = 1 \sum \infty \frac{8}{4 n ^{2} - 1}

4

Solution steps

n = 1 \sum \infty \frac{8}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

\frac{1}{2}

= 8 \cdot \frac{1}{2}

Simplify

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

= 4

n = 0 \sum \infty \frac{2 ^{n}}{n ^{5}}

n = 1 \sum \infty cos^{n} (15)

n = 1 \sum \infty cos^{n} (10)

n = 0 \sum \infty (1 - \frac{2}{n})^{3 n}

k = 0 \sum \infty (- \frac{4}{7})^{k}

What is the sum from n=1 to infinity of 8/(4n^2-1) ?
The sum from n=1 to infinity of 8/(4n^2-1) is 4