to the right but never advance as far to the right as s1; hence according to the same theorem they approach a limit, U.;

but, according to the third part of the hypothesis of the theorem,

U1 = U2,

hence

or let us say U. That is s, approaches a limit U, continually oscillating from one side of its limit to the other.

It is now required to establish analytically the facts on which the plan of the proof rests. It is required to prove:

where the parentheses according to the second part of the hypothesis are all positive (or zero). Hence, the values of §2, §4, §, . . . continually increase, and the values of 8, 8, 85,

6

.. continually

hence,

or to four places, the value of the series is .4055.

The Limit of Error in an Alternating Series

In calculating the value of an infinite series, it is an important matter to know that the value of the series is correct to a given number of decimal places, say to four places.

In order to determine the value of an alternating series correct to the fourth decimal place, it is not sufficient to know that the series is convergent, and that therefore enough terms can be taken so that their sum, 8, will differ from the limit U of the series by less than .0001, since the series might converge so slowly that it would be necessary to take n = 100,000 or greater, so that it would be practically impossible to compute so many terms.

RULE.-The sum of the first n terms of an alternating series, (5), 8659, 8, differ from U, the value of the series, by less than the (n+1)th term. Hence we can stop computing terms as soon as a term is reached which is numerically less than the proposed limit of error.

The proof of this rule follows from the discussion connected with the figure on page 639. 8+1 is determined from s1 by adding ± “m a quantity which is greater than the distance from s, to U. But u is the (n+1)th term of the series. This proves the rule.

6. Compute correct to three decimal places the value of

7. Compute the value of the series in Example 3 correct to four decimal places.

be a convergent series of positive and negative terms.

where r is an integer which is a constant or varies with n.

Proof-Let Sn = U

.

+"n-12

and plot the points 8, 82, 83, Now, when we say that the u series is convergent we mean that s, approaches a limit U; that is, that there is a point U about which the values of s arrange themselves as n increases. In all the series thus far discussed s always came nearer to U as n increased; this is not necessarily Ute

U-e

U

required by the hypothesis of this theorem. Thus s, may be farther away from U than s But the hypothesis of this theorem does require that ultimately s, may be made to differ from U by as small

10,000,000'

a quantity as one chooses. Thus, let e be taken at pleasure 1 say) and lay off an interval e extending from U in each direction, U-e and U+e; then for all values of n taken sufficiently large, say n>p, s, will lie within this interval. This can be formulated algebraically as follows:

Having explained what is meant by "s, approaches a limit U," the proof of the theorem is given. The sum

If n>p, the points corresponding to s, and 8+ will lie in the interval bounded by the points U-e and U+e, and the distance between the points s, and is less than 2 e.

n

Sn+r

Hence

which depends upon n, can be made to remain numerically as small as one pleases by increasing n, hence it approaches 0 as a limit, when, which proves the proposition.

It should be noticed that the condition lim u = 0, is a necessary condition if the series converges but is not a sufficient condition for convergence. For example, in the harmonic series,

not satisfy the general condition of the theorem; for put r = n,

However, the harmonic series does

Thus we have

and does not therefore converge toward the limit 0.

a new proof of the divergence of the harmonic series. It can be proved, however, that the condition

where r can be taken larger than any assigned quantity, is a sufficient condition for the convergence of the series.

On applying the tests of the theorem proved to the series

That is, the series is convergent as the tests of the theorem require.

be a series consisting of an infinite number of positive and negative terms. Let the positive terms in (i) be denoted by the series

taken in the order in which they occur in the series.

For example, if the u series is

where is the number of terms in s, and d their sum.

When n

increases without limit both i and j increase without limit, and two cases may arise.

+un-1,

+wj-1·

di-Pj;