What is the sum from n=1 to infinity of 3(1/5)^{n-1} ?

The sum from n=1 to infinity of 3(1/5)^{n-1} is converges

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sum from n=1 to infinity of 3(1/5)^{n-1}

n = 1 \sum \infty 3 (\frac{1}{5})^{n - 1}

converges

Solution steps

n = 1 \sum \infty 3 (\frac{1}{5})^{n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (\frac{1}{5})^{n - 1}

Apply Series Ratio Test:

converges

= 3 converges

= converges

f (x) = e^{2 x} - 1

n = 1 \sum \infty - n

x \to 0 lim (\frac{sin ( 3 x ^{2} )}{ln ( x ^{2} + 1 )})

\frac{d}{d x} (\frac{x}{( x ^{2} + y ^{2} ) ^{\frac{3}{2}}})

e^{t a n (θ)}

What is the sum from n=1 to infinity of 3(1/5)^{n-1} ?
The sum from n=1 to infinity of 3(1/5)^{n-1} is converges