What is the sum from n=1 to infinity of 6(5/4)^n ?

The sum from n=1 to infinity of 6(5/4)^n is diverges

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sum from n=1 to infinity of 6(5/4)^n

n = 1 \sum \infty 6 (\frac{5}{4})^{n}

diverges

Solution steps

n = 1 \sum \infty 6 (\frac{5}{4})^{n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty (\frac{5}{4})^{n}

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

= 6 (n = 0 \sum \infty (\frac{5}{4})^{n} - (\frac{5}{4})^{0})

n = 0 \sum \infty (\frac{5}{4})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 0 \sum \infty \frac{( 1 ^{n} )}{n !}

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\int_{1}^{25} x ln (x) d x

What is the sum from n=1 to infinity of 6(5/4)^n ?
The sum from n=1 to infinity of 6(5/4)^n is diverges