What is the sum from n=1 to infinity of 3(7/2)^{n-1} ?

The sum from n=1 to infinity of 3(7/2)^{n-1} is diverges

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sum from n=1 to infinity of 3(7/2)^{n-1}

n = 1 \sum \infty 3 (\frac{7}{2})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty 3 (\frac{7}{2})^{n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (\frac{7}{2})^{n - 1}

Apply Series Ratio Test:

diverges

= 3 diverges

= diverges

x \to 0 + lim (tan (3 x))

\frac{d}{d x} (0.5 x^{3})

\int \frac{x}{2 x - 5} d x

y = \frac{1}{x} (1.1)

What is the sum from n=1 to infinity of 3(7/2)^{n-1} ?
The sum from n=1 to infinity of 3(7/2)^{n-1} is diverges