1 2; the following eight terms together are greater than 8 16 or 1 2; and soon. Hence the series is greater than and is therefore divergent. [Art. 475.]
Case III. Let p <1, or negative.
Each term is now greater than the corresponding term in Case II., therefore the series is divergent.
Hence the series is always divergent except in the case when p is positive and greater than unity.
Ex. Prove that the series is divergent.
Compare the given series with 1 +
Thus if u, and v, denote the nth terms of the given series and the auxiliary series respectively, we have hence Lim u_n v_n = 1, and therefore the two series are both convergent or both divergent But the auxiliary series is divergent, therefore also the given series is divergent.
This completes the solution of Ex. 1. [Art. 476.]
480. Convergency of the Binomial Series. To show that the expansion of (1+ x)^n by the Binomial Theorem is convergent when x <1.
Let u_r, u_{r+1}, represent the rth and (r + 1)th terms of the expansion; then
u_{r+1} u_r = {n-r+1} {r} x .
When r>n+1, this ratio is negative; that is, from this point the terms are alternately positive and negative when x is positive, and always of the same sign when x is negative. Now when r is infinite, Lim u_{r+1} u_r = x numerically; therefore since x <1 the series is convergent if all the terms are of the same sign; and therefore stil more is it convergent when some of the terms are positive and some negative. [Art.472.]
