Breakthrough Russian [Breakthrough Lang. Courses] -

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I ank (log nk)'"k 59 Since formula (62) does not hold for every positive s, there exists a positive eo, such that we have I- lim sup {nk! ° -2eo I}nk > 0. I k-. Obviously we then have lim sup {nk ! 1- 1 ° -BU I } nk = 00 Ic o° and, as a consequence, we certainly have I} = 00. 1 Q -e0 I (70) k-+oo Now it follows from (70) that the inequality k! -n,! 1- 1a -e o I>1 I is satisfied for infinitely many values of k. For those values of k we have therefore 1 nk! eo and from this formula we see that the inequality nk 1 1 __I ° I ank I > (log nk)'nk holds for infinitely many values of k.

Now it follows from (37) and (38) that we have 1 (39) _ l l lim sup I any(n)(x) In < (az) a (az) Q = 1. n-+ 00 00 Consequently the series n=U any(")(x) converges for every finite value of x so that the condition is a sufficient one. Now we shall prove that the condition is also a necessary one. 51 Let the differential operator ¢(D) = n=0 anDn be of such a nature that the numbers an have the property that the expression G(z) in (35) does not define an integral function not exceeding the normal type, 1 less than (a-r) a, of the order 1.

2eo I}nk > 0. I k-. Obviously we then have lim sup {nk ! 1- 1 ° -BU I } nk = 00 Ic o° and, as a consequence, we certainly have I} = 00. 1 Q -e0 I (70) k-+oo Now it follows from (70) that the inequality k! -n,! 1- 1a -e o I>1 I is satisfied for infinitely many values of k. For those values of k we have therefore 1 nk! eo and from this formula we see that the inequality nk 1 1 __I ° I ank I > (log nk)'nk holds for infinitely many values of k. From this it follows, in connection with (P, 15), that the series (69) and therefore also the series (68) diverges.

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