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

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

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

n = 1 \sum \infty sin (nπ)

0

Solution steps

n = 1 \sum \infty sin (nπ)

Use the basic trigonometric identity:

sin (nπ) = 0

= n = 1 \sum \infty 0

\sum 0 = 0

= 0

\int_{- 1}^{1} \frac{e ^{2 x}}{1 + e ^{x}} d x

\frac{d}{d x} (\frac{1}{arccos ( x ^{2} + 2 x + 4 )})

f (x) = x^{2}, (- 0.8, 0.64)

\int \frac{5 x}{( x ^{2} + 6 ) ^{2}} d x

\int 0 e^{x} d x

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