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

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

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

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

diverges

Solution steps

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

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= 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 \to \infty lim (5 e^{- t} i - \frac{5 t}{t + 7} j + 5 arctan (t k))

y^{^{'}} = - 6 y

\frac{\partial}{\partial x} (x^{2} + y^{2} + 3 x y + 5 x)

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

\frac{\partial}{\partial y} (2 x e^{- y})

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