Given integers a, b and c, I am trying to find n<=b−1 such that this inequality holds : sum_(i=0)^(n)(a)/(b - i) <= c <= sum_(i=0)^(n+1)(a)/(b-i) That is to say : the largest n up to which the sum is the closest to c . Since : sum_(i=0)^(n)(a)/(b-i)=a*(sum_(i=0)^(b)(1)/(i)-sum_(i=0)^(b-n-1)(1)/(i)) And sum_(i=0)^k 1/i=H_k where H_k is the k-th harmonic number, We can rewrite the inequality into : a(H_b−H_(b−n−1))<=c<=a(H_b−H_(b−n−2)) How to find a tight lower and upper bound for n, such that this inequality holds ?

Aleah Booth 2022-07-17 Answered
Inequality involving sum with parameter in denominator (leading to harmonic number problem)
Given integers a, b and c, I am trying to find n b 1 such that this inequality holds :
i = 0 n a b i c i = 0 n + 1 a b i
That is to say : the largest n up to which the sum is the closest to c
Since :
i = 0 n a b i = a ( i = 0 b 1 i i = 0 b n 1 1 i )
And i = 0 k 1 i = H k where H k is the k-th harmonic number,
We can rewrite the inequality into :
a ( H b H b n 1 ) c a ( H b H b n 2 )
How to find a tight lower and upper bound for n, such that this inequality holds ?
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Answers (1)

agyalapi60
Answered 2022-07-18 Author has 17 answers
Let us rewrite the inequality as
S n c a S n + 1 ,
where S n = 1 b i = 0 n f ( i b ) and f is defined by x 1 1 x for x in [ 0 , 1 [. Thus, one can view S n as the rectangle method of numerical integration applied to f. Since f is strictly increasing, one has
1 b f ( i b ) < i / b ( i + 1 ) / b f ( x ) d x < 1 b f ( i + 1 b ) .
By summation over i, one obtains for all n
S n < 0 ( n + 1 ) / b f ( x ) d x I n + 1 < S n + 1 S 0 < S n + 1 ,
where the integral is given by I n + 1 = ln ( 1 n + 1 b ). Let us examine two cases:
if c / a is larger than S b 1 , then no such n can be found. Taking n = b 1 insures S n c / a only.
else, we can find n such that the first inequality is satisfied. If S n c / a < I n + 1 , then I n < c / a < I n + 1 . Therefore,
n = b ( 1 exp ( c a ) ) ,
where denotes the floor function. Otherwise, I n + 1 c / a S n + 1 , and
n = b ( 1 exp ( c a ) ) 1 .

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