What is the sum from k=3 to infinity of 4/k-4/(k+1) ?

The sum from k=3 to infinity of 4/k-4/(k+1) is 4/3

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sum from k=3 to infinity of 4/k-4/(k+1)

k = 3 \sum \infty \frac{4}{k} - \frac{4}{k + 1}

\frac{4}{3}

Solution steps

k = 3 \sum \infty \frac{4}{k} - \frac{4}{k + 1}

Expand

\frac{4}{k} - \frac{4}{k + 1} : \frac{4}{k ^{2} + k}

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{3}

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

Simplify

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

= \frac{4}{3}

n = 1 \sum \infty \frac{e ^{n}}{π}

n = 0 \sum \infty \frac{1}{n ^{5} + 1}

k = 1 \sum \infty \frac{1}{k ^{5}}

n = 1 \sum \infty \frac{5}{5 n - 1}

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

What is the sum from k=3 to infinity of 4/k-4/(k+1) ?
The sum from k=3 to infinity of 4/k-4/(k+1) is 4/3