What is the sum from n=1 to infinity of ((-1)^n)/3 ?

The sum from n=1 to infinity of ((-1)^n)/3 is diverges

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sum from n=1 to infinity of ((-1)^n)/3

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{3} \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}

= \frac{1}{3} (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 cos (\frac{16}{n})

n = 1 \sum \infty \frac{1}{n ^{4} + 2 n + 8}

n = 0 \sum \infty \frac{n}{n + 4}

n = 0 \sum \infty \frac{3}{n ( n + 2 )}

n = 1 \sum \infty \frac{7}{n ^{2} + 64}

What is the sum from n=1 to infinity of ((-1)^n)/3 ?
The sum from n=1 to infinity of ((-1)^n)/3 is diverges