What is the sum from n=1 to infinity of 3n^5 ?

The sum from n=1 to infinity of 3n^5 is diverges

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sum from n=1 to infinity of 3n^5

n = 1 \sum \infty 3 n^{5}

diverges

Solution steps

n = 1 \sum \infty 3 n^{5}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty n^{5}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

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

\frac{\partial}{\partial z} (3 x sin (y - z))

y^{^{''}} + 5 y^{^{'}} - 14 y = e^{(4 x)}

(- 1.2) (3.3)

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

What is the sum from n=1 to infinity of 3n^5 ?
The sum from n=1 to infinity of 3n^5 is diverges