What is the sum from n=1 to infinity of 4(2.1)^{n-1} ?

The sum from n=1 to infinity of 4(2.1)^{n-1} is diverges

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sum from n=1 to infinity of 4(2.1)^{n-1}

n = 1 \sum \infty 4 (2.1)^{n - 1}

diverges

Solution steps

n = 1 \sum \infty 4 \cdot 2. 1^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 2. 1^{n - 1}

Apply Series Ratio Test:

diverges

= 4 diverges

= diverges

\frac{d}{d x} (3 x^{2} 4^{- x})

t \frac{d y}{d t} = t^{3} + 8 t^{3} y

\int 9 cos (θ) d θ

\frac{d}{d x} (- 2 x^{2} + 7 x + 5)

n = 0 \sum \infty ln (1 + n)

What is the sum from n=1 to infinity of 4(2.1)^{n-1} ?
The sum from n=1 to infinity of 4(2.1)^{n-1} is diverges