What is the sum from n=0 to infinity of i/(n(n+1)) ?

The sum from n=0 to infinity of i/(n(n+1)) is i

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sum from n=0 to infinity of i/(n(n+1))

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

i

Solution steps

i = 1 \sum \infty \frac{n}{i ( i + 1 )}

Apply the constant multiplication rule:

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

= i \cdot n = 1 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

1

= i \cdot 1

Simplify

= i

n = 1 \sum \infty \frac{5 ^{n}}{n ^{3}}

n = 1 \sum \infty \frac{6 e ^{\frac{1}{n}}}{n}

k = 1 \sum \infty 1

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

n = 1 \sum \infty \frac{n}{5 n - 3}

What is the sum from n=0 to infinity of i/(n(n+1)) ?
The sum from n=0 to infinity of i/(n(n+1)) is i