What is the sum from n=1 to infinity of 4^n ?

The sum from n=1 to infinity of 4^n is diverges

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sum from n=1 to infinity of 4^n

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

diverges

Solution steps

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

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

= n = 0 \sum \infty 4^{n} - 4^{0}

n = 0 \sum \infty 4^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

y = 4 x^{2}, (2, 16)

\int_{1}^{10} \frac{1}{x - 6} d x

f (x) = 5 x^{2}, at x = - 3, - 45

f (x) = x^{3} - 4 x

\frac{d}{d x} (\frac{sin ( x ^{2} )}{3 x})

What is the sum from n=1 to infinity of 4^n ?
The sum from n=1 to infinity of 4^n is diverges