What is the sum from n=1 to infinity of 1/(16n^2-8n-3) ?

The sum from n=1 to infinity of 1/(16n^2-8n-3) is 1/4

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sum from n=1 to infinity of 1/(16n^2-8n-3)

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

\frac{1}{4}

Solution steps

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

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

n = 1 \sum \infty \frac{5}{( - 4 ) ^{n}}

n = 1 \sum \infty \frac{n !}{11 4 ^{n}}

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 1 \sum \infty n + 1

What is the sum from n=1 to infinity of 1/(16n^2-8n-3) ?
The sum from n=1 to infinity of 1/(16n^2-8n-3) is 1/4