What is the sum from n=1 to infinity of (4^n)/(3^n) ?

The sum from n=1 to infinity of (4^n)/(3^n) is diverges

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sum from n=1 to infinity of (4^n)/(3^n)

n = 1 \sum \infty \frac{4 ^{n}}{3 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{4 ^{n}}{3 ^{n}}

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

= n = 1 \sum \infty (\frac{4}{3})^{n}

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

= n = 0 \sum \infty (\frac{4}{3})^{n} - (\frac{4}{3})^{0}

n = 0 \sum \infty (\frac{4}{3})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

n = 1 \sum \infty \frac{n 2 ^{n}}{3 ^{n}}

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

n = 1 \sum \infty \frac{n + 5}{n ^{2}}

n = 1 \sum \infty \frac{7 ^{n}}{n}

What is the sum from n=1 to infinity of (4^n)/(3^n) ?
The sum from n=1 to infinity of (4^n)/(3^n) is diverges