What is the sum from n=1 to infinity of 45(-1)^n ?

The sum from n=1 to infinity of 45(-1)^n is diverges

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sum from n=1 to infinity of 45(-1)^n

n = 1 \sum \infty 45 (- 1)^{n}

diverges

Solution steps

n = 1 \sum \infty 45 (- 1)^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 45 \cdot n = 1 \sum \infty (- 1)^{n}

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

= 45 (n = 0 \sum \infty (- 1)^{n} - (- 1)^{0})

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{n ^{6}}{n ^{7}}

n = 1 \sum \infty \frac{4 n}{3 n + 1}

n = 1 \sum \infty \frac{n + 1}{3 - 2 n}

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

n = 0 \sum \infty \frac{i}{n ^{2}}

What is the sum from n=1 to infinity of 45(-1)^n ?
The sum from n=1 to infinity of 45(-1)^n is diverges