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

The sum from n=1 to infinity of 5(1/5)^{n-1} is converges

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sum from n=1 to infinity of 5(1/5)^{n-1}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 1 \sum \infty \frac{ln ( n )}{9 n}

n = 2 \sum \infty \frac{2}{7 ^{n}}

n = 1 \sum \infty \frac{ln ( n )}{3 n}

n = 0 \sum \infty \frac{n}{ln ( n ^{2} )}

n = 1 \sum \infty n^{- s i n (1)}

What is the sum from n=1 to infinity of 5(1/5)^{n-1} ?
The sum from n=1 to infinity of 5(1/5)^{n-1} is converges