What is the sum from n=6 to infinity of (3n)/(12n) ?

The sum from n=6 to infinity of (3n)/(12n) is diverges

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sum from n=6 to infinity of (3n)/(12n)

n = 6 \sum \infty \frac{3 n}{12 n}

diverges

Solution steps

n = 6 \sum \infty \frac{3 n}{12 n}

Apply the constant multiplication rule:

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

= \frac{3}{12} \cdot n = 6 \sum \infty \frac{n}{n}

Cancel the common factor:

3

= \frac{1}{4} \cdot n = 6 \sum \infty \frac{n}{n}

Simplify

= \frac{1}{4} \cdot n = 6 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= \frac{1}{4} diverges

= diverges

y^{^{'''}} + y^{^{'}} = 2 x^{2}

f (x) = e^{t a n (x)}

\frac{\partial}{\partial u} (\frac{v}{1 + uv w})

g (t) = - 4 t^{2} + 3 t - 3

What is the sum from n=6 to infinity of (3n)/(12n) ?
The sum from n=6 to infinity of (3n)/(12n) is diverges