What is the sum from n=1 to infinity of (sin(npi))/n ?

The sum from n=1 to infinity of (sin(npi))/n is 0

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sum from n=1 to infinity of (sin(npi))/n

n = 1 \sum \infty \frac{sin ( nπ )}{n}

0

Solution steps

n = 1 \sum \infty \frac{sin ( nπ )}{n}

\frac{sin ( nπ )}{n} = 0

= n = 1 \sum \infty 0

\sum 0 = 0

= 0

x = 0 \sum \infty \frac{x}{2 ^{x}}

x = 0 \sum \infty \frac{1}{x !}

n = 4 \sum \infty \frac{2}{n ( n - 1 )}

n = 1 \sum \infty \frac{n}{n ^{2} + 7}

n = 0 \sum \infty \frac{1}{n \cdot 3 ^{n}}

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