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

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

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sum from n=1 to infinity of (1)^n

n = 1 \sum \infty (1)^{n}

diverges

Solution steps

n = 1 \sum \infty 1^{n}

Simplify

= n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

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

\int \frac{4 x ^{2} + 7 x + 4}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{x ^{3} - 5 x ^{2}}{x ^{8} - 4})

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

(2 t + 4) i + sin (4 t) j + 2 t^{3} k

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