What is the sum from n=1 to infinity of 6/((-5)^n) ?

The sum from n=1 to infinity of 6/((-5)^n) is converges

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sum from n=1 to infinity of 6/((-5)^n)

n = 1 \sum \infty \frac{6}{( - 5 ) ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{6}{( - 5 ) ^{n}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{( - 5 ) ^{n}}

Apply Series Ratio Test:

converges

= 6 converges

= converges

y^{^{''}} + 256 y = sec (16 x), y (0) = - 5, y^{^{'}} (0) = - 6

\int \frac{cos ( 5 x )}{3 + sin ( 5 x )} d x

y^{^{''}} + 25 y = 5 (csc (5 t))^{2}

t y^{^{'}} + (t + 1) y = t, y (ln (6)) = 1

\frac{\partial}{\partial x} (3 x^{2} + 7 y^{2})

What is the sum from n=1 to infinity of 6/((-5)^n) ?
The sum from n=1 to infinity of 6/((-5)^n) is converges