Page 77 - Foundations Of Differential Calculus

60 3. On the Infinite and the Infinitely Small
106. In order that this may be clarified, let us examine the development
of the fraction 1/ (1 − x) in the first finite number of terms. Hence we have
1 x
=1 + ,
1 − x 1 − x
1 x 2
=1 + x + ,
1 − x 1 − x
3
1 2 x
=1 + x + x + ,
1 − x 1 − x
1 2 3 x 4
=1 + x + x + x + ,
1 − x 1 − x
3
2
and so forth. If someone wishes to say that the finite series 1+x+x +x has
4
a sum equal to 1/ (1 − x), then he is in error by the quantity x / (1 − x);
2 3 1000
if he should say that the sum of the series 1 + x + x + x + ··· + x
1001
is 1/ (1 − x), then his error is equal to x / (1 − x). If x happens to be
greater than 1, this error is very large.
107. From this we see that he who would say that when this same series
is continued to infinity, that is,
3
2
1+ x + x + x + ··· + x ,
∞
∞+1
and that the sum is 1/ (1 − x), then his error would be x / (1 − x), and
if x> 1, then the error is indeed infinite. At the same time, however, this
2 3 4
same argument shows why the series 1 + x + x + x + x + ··· , continued
to infinity, has a true sum of 1/ (1 − x), provided that x is a fraction less
than 1. In this case the error x ∞+1 is infinitely small and hence equal to
1
zero, so that it can safely be neglected. Thus if we let x = , then in truth
2
1 1 1 1 1
1+ + + + + ··· = 1 =2.
2 4 8 16 1 −
2
In a similar way, the rest of the series in which x is a fraction less than 1
will have a true sum in the way we have indicated.
108. This same answer is valid for the sum of divergent series in which
the signs alternate between + and −, which ordinarily is given by the same
formula, but with the sign of x changed to negative. Since we have

1 2 3 4 5
=1 − x + x − x + x − x + ··· ,
1+ x
if we did not express the final remainder, we would have
1
A. 1 − 1+1 − 1+1 − 1+ ··· = ,
2