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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

f (x) = \frac{sin ( x )}{x ^{2}}

y = (\frac{5}{x} + x) (\frac{5}{x} - x)

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

\int \frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} d y

y^{^{'}} + (1 + 3 x^{2}) y = 3 + 9 x^{2}

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