What is the sum from n=1 to infinity of sin(1/(n^2)) ?

The sum from n=1 to infinity of sin(1/(n^2)) is converges

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sum from n=1 to infinity of sin(1/(n^2))

n = 1 \sum \infty sin (\frac{1}{n ^{2}})

converges

Solution steps

n = 1 \sum \infty sin (\frac{1}{n ^{2}})

Apply Series Comparison Test:

converges

= converges

\int - \frac{1}{sin ( θ )} d θ

y^{^{'}} + t^{2} y = t^{2}

\frac{d}{d x} (9 x e^{- \frac{1}{y ^{3}}})

\int \frac{1}{6 - x ^{2}} d x

\frac{d}{d x} (\frac{5}{7 x ^{2}})

What is the sum from n=1 to infinity of sin(1/(n^2)) ?
The sum from n=1 to infinity of sin(1/(n^2)) is converges