What is the sum from n=1 to infinity of ((-1)^n)/(5(2^{n+2))} ?

The sum from n=1 to infinity of ((-1)^n)/(5(2^{n+2))} is converges

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sum from n=1 to infinity of ((-1)^n)/(5(2^{n+2))}

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{5 ( 2 ^{n + 2} )}

converges

Solution steps

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{5 \cdot 2 ^{n + 2}}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n + 2}}

Simplify

2^{n + 2} : 2^{n} \cdot 2^{2}

= \frac{1}{5} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n} \cdot 2 ^{2}}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot \frac{1}{2 ^{2}} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n}}

Simplify

= \frac{1}{5} \cdot \frac{1}{4} \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n}}{2 ^{n}}

Apply Alternating Series Test:

converges

= \frac{1}{5} \cdot \frac{1}{4} converges

= converges

x \to 1 + lim (\frac{7}{1 - x})

y = 4 - 3 x

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\int 18 x^{2} 6 x^{3} + 5 d x

\frac{\partial}{\partial y} (6 y^{2})

What is the sum from n=1 to infinity of ((-1)^n)/(5(2^{n+2))} ?
The sum from n=1 to infinity of ((-1)^n)/(5(2^{n+2))} is converges