What is the sum from n=1 to infinity of (2^n)/(100) ?

The sum from n=1 to infinity of (2^n)/(100) is diverges

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sum from n=1 to infinity of (2^n)/(100)

n = 1 \sum \infty \frac{2 ^{n}}{100}

diverges

Solution steps

n = 1 \sum \infty \frac{2 ^{n}}{100}

Apply the constant multiplication rule:

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

= \frac{1}{100} \cdot n = 1 \sum \infty 2^{n}

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

= \frac{1}{100} (n = 0 \sum \infty 2^{n} - 2^{0})

n = 0 \sum \infty 2^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int x^{2} cos (x + 1) d x

\frac{\partial}{\partial y} (- y sin (x y))

\int cos (2 a x) d x

\frac{d}{d x} ((5 x^{2} + e)^{6})

n = 3 \sum \infty x^{n}

What is the sum from n=1 to infinity of (2^n)/(100) ?
The sum from n=1 to infinity of (2^n)/(100) is diverges