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

The sum from n=1 to infinity of 1+sin^2(n) is diverges

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 5 \sum \infty (\frac{2}{11})^{n}

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

k = 1 \sum \infty \frac{k - 5}{k + 2}

n = 0 \sum \infty \frac{1}{n \cdot ( n + 1 )}

n = 1 \sum \infty (\frac{- 1}{3})^{n} n

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