What is the sum from n=0 to infinity of cos(2npi) ?

The sum from n=0 to infinity of cos(2npi) is diverges

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sum from n=0 to infinity of cos(2npi)

n = 0 \sum \infty cos (2 nπ)

diverges

Solution steps

n = 0 \sum \infty cos (2 nπ)

Rewrite using trig identities

= n = 0 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

n = 0 \sum \infty \frac{15}{4} (\frac{1}{4})^{n}

n = 1 \sum \infty \frac{1}{2 ^{n} - 9}

n = 7 \sum \infty \frac{1}{n ( n + 1 )}

n = 1 \sum \infty \frac{7}{n ^{2} + 16}

n = 0 \sum \infty \frac{n}{7 n ^{3} + 1}

What is the sum from n=0 to infinity of cos(2npi) ?
The sum from n=0 to infinity of cos(2npi) is diverges