What is the sum from n=1 to infinity of 3*(4/3)^n ?

The sum from n=1 to infinity of 3*(4/3)^n is diverges

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sum from n=1 to infinity of 3*(4/3)^n

n = 1 \sum \infty 3 \cdot (\frac{4}{3})^{n}

diverges

Solution steps

n = 1 \sum \infty 3 (\frac{4}{3})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (\frac{4}{3})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 3 (n = 0 \sum \infty (\frac{4}{3})^{n} - (\frac{4}{3})^{0})

n = 0 \sum \infty (\frac{4}{3})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- 2 t}

x (81.9)

(1 + x)^{2} y^{^{'}} + (1 + x) y = 1

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

(t - \frac{3 π}{6}) sin (6 t)

What is the sum from n=1 to infinity of 3*(4/3)^n ?
The sum from n=1 to infinity of 3*(4/3)^n is diverges