What is the sum from n=1 to infinity of (7/6)^n ?

The sum from n=1 to infinity of (7/6)^n is diverges

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sum from n=1 to infinity of (7/6)^n

n = 1 \sum \infty (\frac{7}{6})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{7}{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 (\frac{7}{6})^{n} - (\frac{7}{6})^{0}

n = 0 \sum \infty (\frac{7}{6})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x y - 2 y + x y^{^{'}} - y^{^{'}} = 0

(4, - 1), (- 9, - 10)

sin^{3} (x^{2})

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

x \to 49 lim (x^{\frac{3}{2}})

What is the sum from n=1 to infinity of (7/6)^n ?
The sum from n=1 to infinity of (7/6)^n is diverges