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

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8 y^{^{''}} - 7 y^{^{'}} + 3 y = 0

\int 5 x e^{- x^{2}} d x

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