please; that is, the series 1 + r + rr + &c. continually approaches to the limit
| 1 | . |
|---|---|
| 1 - r |
Thus 1 + ½ + ¼ + ⅛ + &c. where r = ½, continually approaches to
| 1 | or 2 , |
|---|---|
| 1 - ½ |
as was shewn in the last article.
EXERCISES.
| The limit of | 2 | + | 2 | + | 2 | + &c. |
|---|---|---|---|---|---|---|
| 3 | 9 |
| or | 2 | ( | 1 | + | 1 | + | 1 | + &c. | ) | is 3 |
|---|---|---|---|---|---|---|---|---|---|---|
| 3 | 9 |
| ... | 1 | + | 9 | + | 81 | + &c. | ... 10 |
|---|---|---|---|---|---|---|---|
| 10 | 100 | ||||||
| ... | 5 | + | 15 | + | 45 | + &c. | ... 8¾ |
| 7 | 49 |
- When the fraction a/b is not equal to c/d, but greater, a is said to have to b a greater ratio than c has to d; and when a/b is less than c/d, a is said to have to b a less ratio than c has to d. We propose the following questions as exercises, since they follow very simply from this definition.
I. If a be greater than b, and c less than or equal to d, a will have a greater ratio to b than c has to d.
II. If a be less than b, and c greater than or equal to d, a has a less ratio to b than c has to d.
III. If a be to b as c is to d, and if a have a greater ratio to b than c has to x, d is less than x; and if a have a less ratio to b than c to x, d is greater than x.
IV. a has to b a greater ratio than ax to bx + y, and a less ratio than ax to bx- y.
- If a have to b a greater ratio than c has to d, a + c has to b + d a less ratio than a has to b, but a greater ratio than c has to d; or, in other words, if a/b be the greater of the two fractions a/b and c/d,
| a + c |
|---|
| b + d |
will be greater than c / d , but less than a / b . To shew this, observe that ( mx + ny )/( m + n ) must lie between x and y