What is the sum from k=1 to infinity of 8(1/5)^k ?

The sum from k=1 to infinity of 8(1/5)^k is 2

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sum from k=1 to infinity of 8(1/5)^k

k = 1 \sum \infty 8 (\frac{1}{5})^{k}

2

Solution steps

k = 1 \sum \infty 8 (\frac{1}{5})^{k}

Apply the constant multiplication rule:

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

= 8 \cdot k = 1 \sum \infty (\frac{1}{5})^{k}

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

= 8 (k = 0 \sum \infty (\frac{1}{5})^{k} - (\frac{1}{5})^{0})

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

= 8 (\frac{5}{4} - (\frac{1}{5})^{0})

Simplify

8 (\frac{5}{4} - (\frac{1}{5})^{0}) : 2

= 2

sin (x) cos (x)

\frac{8 x}{( x ^{2} + 1 )}

x \to 0 - lim (ln (x))

\int \frac{8}{( 4 - t ) ^{8}} d t

y = 8 + 4 x^{2} - 2 x^{3}, (2, 8)

What is the sum from k=1 to infinity of 8(1/5)^k ?
The sum from k=1 to infinity of 8(1/5)^k is 2