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

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

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

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

diverges

Solution steps

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (- 1)^{n} - (- 1)^{0} - (- 1)^{1}

n = 0 \sum \infty (- 1)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 0 + lim (\frac{36 - 4 x ^{2}}{x})

\int_{0}^{\infty} \frac{1}{2 + e ^{x}} d x

\int \frac{1}{8 - 5 x} d x

+ o f + \frac{x - 4}{x + 4}

\int \frac{x ^{5} - 3125}{x - 5} d x

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