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

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

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sum from n=1 to infinity of 3/(4n^2+4n)

n = 1 \sum \infty \frac{3}{4 n ^{2} + 4 n}

\frac{3}{4}

Solution steps

n = 1 \sum \infty \frac{3}{4 n ^{2} + 4 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{4 n ^{2} + 4 n}

Apply Telescoping Series Test:

\frac{1}{4}

= 3 \cdot \frac{1}{4}

Simplify

3 \cdot \frac{1}{4} : \frac{3}{4}

= \frac{3}{4}

k = 1 \sum \infty \frac{2 ^{k}}{k ^{3}}

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

n = 0 \sum \infty \frac{6}{n ^{3}}

k = 0 \sum \infty (\frac{1}{8})^{k}

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

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