What is the sum from n=1 to infinity of (5^n)/9 ?

The sum from n=1 to infinity of (5^n)/9 is diverges

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sum from n=1 to infinity of (5^n)/9

n = 1 \sum \infty \frac{5 ^{n}}{9}

diverges

Solution steps

n = 1 \sum \infty \frac{5 ^{n}}{9}

Apply the constant multiplication rule:

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

= \frac{1}{9} \cdot n = 1 \sum \infty 5^{n}

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

= \frac{1}{9} (n = 0 \sum \infty 5^{n} - 5^{0})

n = 0 \sum \infty 5^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

f (x) = \frac{sin ( 4 x )}{sin ( 5 x )}

y = arctan (3 x^{2})

\int e^{10 t} d t

\frac{d}{d x} (f (x)^{- 1} (x))

x^{2} y^{^{'}} + 2 x y = 21 cos^{2} (x)

What is the sum from n=1 to infinity of (5^n)/9 ?
The sum from n=1 to infinity of (5^n)/9 is diverges