sum from n=0 to infinity of 4(3/4)^n

Let's calculate:

\sum_{n = 0}^{+∞} 4 \cdot {(\frac{3}{4})}^{n}

The series is geometric, with first term

4

and ratio

\frac{3}{4}

. Since

| \frac{3}{4} | < 1

, its sum is

\frac{4}{1 - \frac{3}{4}}

ⅰ

\sum_{n = 0}^{+∞} 4 \cdot {(\frac{3}{4})}^{n}

\frac{4}{1 - \frac{3}{4}}

\frac{4}{1 - \frac{3}{4}}

ⓘ To add fractions, put them over the same denominator, then add the numerators:

1 - \frac{3}{4}

\frac{4 - 3}{4}

\frac{4}{\frac{4 - 3}{4}}

ⓘ The numerator and the denominator are computed before the division:

4 - 3

1

\frac{4}{\frac{1}{4}}

ⓘ Dividing is multiplying by the reciprocal:

\frac{4}{\frac{1}{4}}

4 \cdot 4

4 \cdot 4

▶Result:

16