What is the sum from n=1 to infinity of (-9^n)/(n^4) ?

The sum from n=1 to infinity of (-9^n)/(n^4) is diverges

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sum from n=1 to infinity of (-9^n)/(n^4)

n = 1 \sum \infty \frac{- 9 ^{n}}{n ^{4}}

diverges

Solution steps

n = 1 \sum \infty \frac{- 9 ^{n}}{n ^{4}}

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty \frac{9 ^{n}}{n ^{4}}

n \to \infty lim (n (\frac{\frac{9 ^{n}}{n ^{4}}}{\frac{9 ^{n + 1}}{( n + 1 ) ^{4}}} - 1)) = - \infty

L < 1,

by the Raabe's test

= - diverges

= diverges

\int (x^{2} + 2 x) \cdot 3^{x} d x

\int_{- 3}^{3} (13 - x^{2}) - (x^{2} - 5) d x

\int \frac{1}{x ^{2} + 169} d x

(2 x + 1) (\frac{d y}{d x}) + y = (2 x + 1)^{\frac{3}{2}}

y^{^{''}} + 3 y = - 48 x^{2} e^{3 x}

What is the sum from n=1 to infinity of (-9^n)/(n^4) ?
The sum from n=1 to infinity of (-9^n)/(n^4) is diverges