sum(-3*(1/2)^(2*n+1),n=1,+inf)

Let's evaluate:

\sum_{n = 1}^{+∞} -3 \cdot {(\frac{1}{2})}^{2 n + 1}

The series is geometric, with first term

- \frac{3}{8}

and ratio

\frac{1}{4}

. Since

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

, its sum is

- \frac{\frac{3}{8}}{1 - \frac{1}{4}}

ⅰ

\sum_{n = 1}^{+∞} - 3 \cdot {(\frac{1}{2})}^{2 n + 1}

- \frac{\frac{3}{8}}{1 - \frac{1}{4}}

- \frac{\frac{3}{8}}{1 - \frac{1}{4}}

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

1 - \frac{1}{4}

\frac{4 - 1}{4}

- \frac{\frac{3}{8}}{\frac{4 - 1}{4}}

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

4 - 1

3

- \frac{\frac{3}{8}}{\frac{3}{4}}

ⓘ Dividing is multiplying by the reciprocal:

- \frac{\frac{3}{8}}{\frac{3}{4}}

- \frac{3}{8} \cdot \frac{4}{3}

- \frac{3}{8} \cdot \frac{4}{3}

ⓘ To multiply fractions, multiply the numerators together and the denominators together:

- \frac{3}{8} \cdot \frac{4}{3}

- \frac{3 \cdot 4}{8 \cdot 3}

- \frac{12}{24}

- \frac{12}{24}

The fraction:

\frac{12}{24}

simplifies by the common factor:

12

into the new quotient:

\frac{1}{2}

▶Result:

- \frac{1}{2}

The numeric value of:

- \frac{1}{2}

is:

▷

-0.5