An infinite series $\displaystyle \sum_{i = 1}^{\infty}a^{i}$ converges to a sum if and only if $\displaystyle |a|< 1$. However, in the series $\displaystyle \sum_{i = 1}^{\infty} \left (- \frac{5}{4} \right )^{i}$, this is not the case, as $\displaystyle \left | - \frac{5}{4} \right |= \frac{5}{4} > 1$. This series diverges.

The sum of an infinite series $\displaystyle \sum_{i = 1}^{\infty}a^{i}$, where $\displaystyle |a| < 1$, can be calculated as follows:

$\displaystyle \sum_{i = 1}^{\infty} a^{i} = \frac{a }{1- a}$

Setting $\displaystyle a = \frac{4}{7}$:

$\displaystyle \sum_{i = 1}^{\infty} \left (\frac{4}{7} \right )^{i} = \frac{\frac{4}{7}}{1-\frac{4}{7}}= \frac{\frac{4}{7}}{ \frac{3}{7} } = \frac{4}{7} \div \frac{3}{7} = \frac{4}{7} \cdot \frac{7} {3} = \frac{4}{3}$

Write the formula for infinite geometric series.

$\displaystyle S=\frac{a_1}{1-r}$

The value of $\displaystyle a_1$ is the first term of the series, which is $\displaystyle \frac{3}{4}$.

The value of the common ratio, $\displaystyle r$, is also $\displaystyle \frac{3}{4}$.

The ratio is $\displaystyle \frac{3}{4}$ because if we were to write out the first few terms in the series we would see,

$\displaystyle \frac{3}{4}^1+\frac{3}{4}^2+ \frac{3}{4}^3+...$

each term is three fourths more than the previous term therefore, giving us the ratio.

Substitute the values into the equation and evaluate.

$\displaystyle S=\frac{\frac{3}{4}}{1-\frac{3}{4}} = \frac{\frac{3}{4}}{\frac{1}{4}} =\frac{3}{4}\cdot4 = 3$

Write the infinite series formula.

$\displaystyle S=\frac{a_1}{1-r}$

Substitute the values of $\displaystyle a_1$ and $\displaystyle r$.

$\displaystyle S=\frac{a_1}{1-r}= \frac{5}{1-\frac{1}{4}} = \frac{5}{\frac{3}{4}} = 5\cdot\frac{4}{3}=\frac{20}{3}$

The first term of this sequence is 10. To find the common ratio r, we can just divide the second term by the first: $\displaystyle -9 \div 10 = - 0.9$. So "r" is -0.9. We can find the infinite sum using the formula $\displaystyle \frac{a}{1-r}$ where a is the first term and r is the common ratio: 

$\displaystyle \frac{10}{1-(-0.9)} = \frac{10}{1+0.9 } = \frac{10}{1.9} \approx 5.263$

An infinite sum is only calculable if $\displaystyle \left | r \right | < 1$ where r is the common ratio. We can find the common ratio easily by dividing the second term by the first: $\displaystyle 11 \div 10 = 1.1$. This is greater than 1, so we can't find the infinite sum - it is infinite.

Write the formula to find the sum of an infinite geometric series.

$\displaystyle S=\frac{a_1}{1-r}$

The first term is: $\displaystyle a_1 = \frac{1}{8}$.

The common ratio is:  $\displaystyle r=\frac{1}{2}$

Substitute the values into the formula.

$\displaystyle S=\frac{\frac{1}{8}}{1-\frac{1}{2}} = \frac{\frac{1}{8}}{\frac{1}{2}}$

Rewrite the complex fraction.

$\displaystyle S=\frac{1}{8}\div \frac{1}{2} = \frac{1}{8} \times 2 = \frac{1}{4}$

The sum will converge to $\displaystyle \frac{1}{4}$.

Write the formula for the sum of an infinite series.

$\displaystyle S=\frac{a_1}{1-r}$

The value $\displaystyle a_1$ is the first term, and $\displaystyle r$ is the common ratio.

$\displaystyle a_1=4$

Divide the second term with the first term, third term and the second, and so forth, and we will get a common ratio of:

$\displaystyle r=-\frac{1}{2}$

Substitute the values into the formula.

$\displaystyle S=\frac{a_1}{1-r}= \frac{4}{1-(-\frac{1}{2})} = \frac{4}{\frac{3}{2}}$

Rewrite the complex fraction using a division sign.

$\displaystyle \frac{4}{\frac{3}{2}} = 4 \div \frac{3}{2}$

Change the division sign to a multiplication and take the reciprocal of the second term.

$\displaystyle 4 \times \frac{2}{3} = \frac{8}{3}$

The series will converge to $\displaystyle \frac{8}{3}$.

Write the formula for finding the sum of an infinite geometric series.

$\displaystyle S=\frac{a_1}{1-r}$

The first term $\displaystyle a_1$ is ten.

Find the common ratio by dividing the second term with the first term, third term with the second term, or the fourth term with the third term, and so forth.  

The common ratio should all be the same after dividing each term.

$\displaystyle r=\frac{-6}{10} = -\frac{3}{5}$

As long as $\displaystyle r$ is between negative one and one, we can use the formula to find the sum.  Substitute the givens into the equation and solve.

$\displaystyle S=\frac{10}{1-(-\frac{3}{5})} = \frac{10}{\frac{8}{5}} = 10\times \frac{5}{8}= \frac{50}{8}= \frac{25}{4}$

The answer is:  $\displaystyle \frac{25}{4}$

Write the formula for an infinite series.

$\displaystyle S=\frac{a_1}{1-r}$

 The term $\displaystyle a_1$ represents the first term.  The value of the common ratio $\displaystyle r$ must be between negative one and positive one, and can be determined by dividing the second term with the first term, third term with the second term, and so forth.

$\displaystyle \frac{2}{9}\div 2 = \frac{2}{9} \times \frac{1}{2} =\frac{1}{9}$

$\displaystyle \frac{2}{81}\div\frac{2}{9} = \frac{2}{81}\times\frac{9}{2} = \frac{9}{81} = \frac{1}{9}$

We can see that the common ratio is:  $\displaystyle r=\frac{1}{9}$

Substitute the known values into the formula.

$\displaystyle S=\frac{2}{1-\frac{1}{9}} = \frac{2}{\frac{8}{9}} = 2\div \frac{8}{9} = 2\times \frac{9}{8}= \frac{9}{4}$

The sum will converge to:  $\displaystyle \frac{9}{4}$