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

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

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sum from n=0 to infinity of sin(npi)

n = 0 \sum \infty sin (nπ)

0

Solution steps

n = 0 \sum \infty sin (nπ)

Use the basic trigonometric identity:

sin (nπ) = 0

= n = 0 \sum \infty 0

\sum 0 = 0

= 0

(2 x^{3} - x y^{2} - 2 y + 3) d x - (x^{2} y + 2 x) d y = 0

t^{2} e^{3 t}

\int sin^{2} (4 x) cos^{2} (4 x) d x

\frac{d}{d x} (5^{s i n (π x)})

\frac{e ^{x}}{1}

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