What is the sum from n=1 to infinity of (1.06)^n ?

The sum from n=1 to infinity of (1.06)^n is diverges

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sum from n=1 to infinity of (1.06)^n

n = 1 \sum \infty (1.06)^{n}

diverges

Solution steps

n = 1 \sum \infty 1.0 6^{n}

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

= n = 0 \sum \infty 1.0 6^{n} - 1.0 6^{0}

n = 0 \sum \infty 1.0 6^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to - 2 lim (\frac{2 x ^{2}}{2 x + 2})

x - 5 + \frac{1}{x}

x^{9 x}

y^{^{'}} + 8 (tan (8 x)) \cdot y = - 9 cos (8 x)

(0.5) (7.9)

What is the sum from n=1 to infinity of (1.06)^n ?
The sum from n=1 to infinity of (1.06)^n is diverges