What is the sum from n=0 to infinity of sin(1/(n^2)) ?

The sum from n=0 to infinity of sin(1/(n^2)) is converges

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sum from n=0 to infinity of sin(1/(n^2))

n = 0 \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

x \to 0 lim (\frac{\frac{1}{1 + x} - 1}{x})

ln (3 - x^{3})

\int ln (x^{2} + x + 1) d x

\int 8 sec^{3} (x) d x

\frac{\partial}{\partial y} (x^{2} \cdot ln (y))

What is the sum from n=0 to infinity of sin(1/(n^2)) ?
The sum from n=0 to infinity of sin(1/(n^2)) is converges