What is the sum from n=1 to infinity of ((-5)^n)/4 ?

The sum from n=1 to infinity of ((-5)^n)/4 is diverges

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sum from n=1 to infinity of ((-5)^n)/4

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{4}

diverges

Solution steps

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{4}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 1 \sum \infty (- 5)^{n}

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

= \frac{1}{4} (n = 0 \sum \infty (- 5)^{n} - (- 5)^{0})

n = 0 \sum \infty (- 5)^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

k = 0 \sum \infty (0.1)^{k}

n = 0 \sum \infty \frac{2}{7 - 3 n}

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

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

n = 0 \sum \infty 5 \cdot (\frac{2}{3})^{n}

What is the sum from n=1 to infinity of ((-5)^n)/4 ?
The sum from n=1 to infinity of ((-5)^n)/4 is diverges