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Concrete mathematics : a foundation for computer science

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Hence S, = a7m-l -- inP2 - An3 + O(nP5).<br />

9.37 This is<br />

k,q>l = n2-<br />

= ,2 -.<br />

A ANSWERS TO EXERCISES 571<br />

The remaining sum is like (9.55) but without the factor u(q). The same<br />

method works here as it did there, but we get L(2) in place of l/ 1 is 0 (1). Suppose n is even. Euler’s<br />

summation <strong>for</strong>mula implies that<br />

hence the sum is i H,” + 0 (1). In general the answer is 5 (- -l)nH,m -t O(1).<br />

9.41 Let CX= $/L$ = -@-2. We have<br />

ClnFk = ~(h~k-h&+h(l -ak))<br />

k=l<br />

z<br />

1)<br />

(In eYn)m +0(l)<br />

m<br />

n(n + 1)<br />

In@-5ln5+tln(l -ak)-xln(l -elk).<br />

2<br />

k21 k>n<br />

The latter sum is tIk>,, O(K~) = O(~L~). Hence the answer is<br />

@+1/25-Wc + o&n’” 31/+-n/Z) , where<br />

C = (1 -a)(1 -~~)(l -K~)... zz 1.226742.

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