What is the sum from n=2 to infinity of 5*(-3/4)^n ?

The sum from n=2 to infinity of 5*(-3/4)^n is converges

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sum from n=2 to infinity of 5*(-3/4)^n

n = 2 \sum \infty 5 \cdot (- \frac{3}{4})^{n}

converges

Solution steps

n = 2 \sum \infty 5 (- \frac{3}{4})^{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 2 \sum \infty (- \frac{3}{4})^{n}

Apply Series Ratio Test:

converges

= 5 converges

= converges

y^{^{''}} + 5 y^{^{'}} + 4 y = t^{2} e^{7 t}

\int_{0}^{l n (5)} e^{x} - e^{- 2 x} d x

\int_{0}^{6} \frac{6 t}{( t - 7 ) ^{2}} d t

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

(x + 3) (x - 5)^{7}

What is the sum from n=2 to infinity of 5*(-3/4)^n ?
The sum from n=2 to infinity of 5*(-3/4)^n is converges