What is the sum from n=2 to infinity of 1/(n4^n) ?

The sum from n=2 to infinity of 1/(n4^n) is converges

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sum from n=2 to infinity of 1/(n4^n)

n = 2 \sum \infty \frac{1}{n 4 ^{n}}

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

6 e^{x} cos (x)

f (x) = - \frac{1}{( 5 x + 1 ) ^{2}}

4 y^{^{''}} + y^{^{'}} = 0

x \to 2 lim (\frac{2}{0})

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

What is the sum from n=2 to infinity of 1/(n4^n) ?
The sum from n=2 to infinity of 1/(n4^n) is converges