Approximating Sums of Consecutive Integral Roots

Pongpol Ruankong and Kantaphon Kuhapatanakul Department of Mathematics Faculty of Science Kasetsart University Bangkok 10900 Thailand fscippru@ku.ac.th fscikpkk@ku.ac.th

Abstract We present an alternative proof of Saltzman and Yuan’s result on the sums of consecutive integral roots. We use the AM-GM-HM inequality to prove the main result. Moreover, the lower bound for which the result holds is greatly improved.

1 Introduction

For each real number x, let ⌊x⌋ denote the greatest integer not exceeding x. The sums of consecutive integral roots have been studied by many mathematicians (see [1]-[5]). For instance, the following identities hold for every positive integer n:

(1) ⌊\sqrt{n} + \sqrt{n + 1}⌋ = ⌊\sqrt{4n + 1}⌋

(2) ⌊\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2}⌋ = ⌊\sqrt{9n + 8}⌋

(3) ⌊\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} + \sqrt{n + 3}⌋ = ⌊\sqrt{16n + 20}⌋

(4) ⌊\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} + \sqrt{n + 3} + \sqrt{n + 4}⌋ = ⌊\sqrt{25n + 49}⌋ (5) ⌊\sqrt[3]{n} + \sqrt[3]{n + 1}⌋ = ⌊\sqrt[3]{8n + 3}⌋

(6) ⌊\sqrt[3]{n} + \sqrt[3]{n + 1} + \sqrt[3]{n + 2}⌋ = ⌊\sqrt[3]{27n + 26}⌋.

On the contrary, Zhan [5] showed that for any real number c, there is a positive integer n such that

⌊\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} + \sqrt{n + 3} + \sqrt{n + 4} + \sqrt{n + 5}⌋ ≠ ⌊\sqrt{36n + c}⌋ .

Saltzman and Yuan [3] presented the following similar formula for n ≥ 4.

⌊\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} + \sqrt{n + 3} + \sqrt{n + 4} + \sqrt{n + 5}⌋ = ⌊\sqrt{36n + 89}⌋ ,

and gave the general formula: for all integers p, m ≥ 2,

⌊\sqrt[p]{n} + \sqrt[p]{n + 1} + \sqrt[p]{n + 2} + ⋯ + \sqrt[p]{n + m − 1}⌋ = ⌊\sqrt[p]{m^{p}n + \frac{m^{p}(m − 1)}{2} − 1}⌋ (1)

holds for all positive integers

n > \frac{m^{p}(m − 1)(2m − 1)(p − 1)}{12p}.  (2)

They utilized properties of concave functions to approximate the sums of consecutive integral roots.

In this paper, we give an alternative proof of (1) using the AM-GM-HM inequality and obtain a new lower bound on n, which is approximately half of (2).

2 Results

In the sequel, let m and p be two positive integers greater than 1, let M = \frac{(p−1)m^{p}(m^{2}−1)}{12p}, and let i_{1}, i_{2}, …, i_{p} ∈ {0, 1, 2, …, m − 1}. Denoted by A_{p}, G_{p} and H_{p} the arithmetic mean, the geometric mean, and the harmonic mean of the positive integers n + i_{1}, n + i_{2}, …, n + i_{p}, respectively, i.e.,

A_{p} = \frac{1}{p} ∑_{j=1}^{p}(n + i_{j}),
G_{p} = \sqrt[p]{(n + i_{1}) … (n + i_{p})},
H_{p} = \frac{p}{∑{p \atop j=1} \frac{1}{n+i_{j}}} .

By the well-known AM-GM-HM inequality,

H_{p} ≤ G_{p} ≤ A_{p}

where equality holds if and only if n+i_{1} = n+i_{2} = ⋯ = n+i_{p}. To prove the main theorem, we first present two technical lemmas.

Lemma 1. For every positive integer n,

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (A_{p} − H_{p}) < \frac{M}{n^{p−1}} (n + \frac{m − 1}{2})^{p−2} .

Proof. Observe that

A_{p} − H_{p} = \frac{1}{p} ∑_{j=1}^{p}(n + i_{j}) − \frac{p}{∑{p \atop j=1} \frac{1}{n+i_{j}}}

= \frac{((n + \frac{1}{p} ∑_{j=1}^{p} i_{j}) (\frac{1}{p} ∑_{k=1}^{p} \frac{∏_{j=1}^{p}(n+i_{j})}{n+i_{k}}) − ∏_{j=1}^{p}(n + i_{j}))}{\frac{1}{p} ∑{p \atop k=1} \frac{∏_{j=1}^{p}(n+i_{j})}{n+i_{k}}}.

Hence

\frac{A_{p} − H_{p}}{p} ∑_{k=1}^{p} \frac{∏_{j=1}^{p}(n + i_{j})}{n + i_{k}} = (n + \frac{1}{p} ∑_{j=1}^{p} i_{j}) (\frac{1}{p} ∑_{k=1}^{p} \frac{∏_{j=1}^{p}(n + i_{j})}{n + i_{k}}) − ∏_{j=1}^{p}(n + i_{j}).

Next, we determine the coefficients of

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} ((n + \frac{1}{p} ∑_{j=1}^{p} i_{j}) (\frac{1}{p} ∑_{k=1}^{p} \frac{∏_{j=1}^{p}(n + i_{j})}{n + i_{k}}) − ∏_{j=1}^{p}(n + i_{j}))  (3)

as a polynomial in variable n.

It is easy to see that the coefficients of n^{p} and n^{p−1} in (3) are zero. Moreover, we claim that, for each t = 2, 3, …, p, the coefficient of n^{p−t} in (3) is equal to M\binom{p−2}{t−2} (\frac{m−1}{2})^{t−2} implying that the polynomial is M (n + \frac{m−1}{2})^{p−2}.

The coefficient of n^{p−t} in (3), for t = 2, 3, …, p, is equal to the sums of the following:

(i) ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (\frac{1}{p^{2}} ∑_{j=1}^{p} i_{j} (∑_{S} ∑_{\begin{matrix}j_{1}<···<j_{t−1} \\ j_{1},...,j_{t−1}∈S \end{matrix}} i_{j_{1}} ⋯ i_{j_{t−1}} ))

(ii) ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (\frac{1}{p} ∑_{S} ∑_{\begin{matrix}j_{1}<···<j_{t} \\ j_{1},...,j_{t}∈S \end{matrix}} i_{j_{1}} ⋯ i_{j_{t}} )

(iii) ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (− ∑_{\begin{matrix}j_{1}<···<j_{t} \\ j_{1},...,j_{t}∈ℕ_{p} \end{matrix}} i_{j_{1}} ⋯ i_{j_{t}} )

where ∑_{S} represents the sum over all (p − 1)-subsets S of ℕ_{p} = {1, 2, …, p}. (Note that for the case t = p, (ii) is an empty sum which is conventionally zero. This coincides with the fact that (ii) does not occur in the case t = p.)

Distributing ∑_{j=1}^{p} i_{j} inside the double sum, (i) becomes

\frac{1}{p^{2}} ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (∑_{S} ∑_{\begin{matrix}j_{1}<···<j_{t−1} \\ j_{1},...,j_{t−1}∈S \end{matrix}} (∑_{r=1}^{t−1} i_{j_{r}}(i_{j_{1}} ⋯ i_{j_{t−1}}) + ∑_{r∉\{1,...,t−1\}} i_{j_{r}}(i_{j_{1}} ⋯ i_{j_{t−1}}))) .

Since the double sum consists of exactly p\binom{p−1}{t−1} terms, using the symmetry of the i_{j} and combining the like terms, (i) amounts to

\frac{1}{p}\binom{p − 1}{t − 1} ((t − 1) ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1}^{2} ⋯ i_{t−1} + (p − t + 1) ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1} ⋯ i_{t}) .  (4)

Similarly, (ii) amounts to

\binom{p − 1}{t} ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1} ⋯ i_{t},  (5)

and (iii) amounts to

− \binom{p}{t} ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1} ⋯ i_{t}.  (6)

Note that

\frac{p − t + 1}{p}\binom{p − 1}{t − 1} + \binom{p − 1}{t} − \binom{p}{t} = −\frac{t − 1}{p}\binom{p − 1}{t − 1}

for t = 2, …, p. Hence adding (4), (5) and (6) yields

\frac{t − 1}{p}\binom{p − 1}{t − 1} (∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1}^{2} ⋯ i_{t−1} − ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} i_{1} ⋯ i_{t})

= \frac{p − 1}{p}\binom{p − 2}{t − 2} (m^{p−t+1} ∑_{i=0}^{m−1} i^{2} (∑_{i=0}^{m−1} i)^{t−2} − m^{p−t} (∑_{i=0}^{m−1} i)^{t})

= \frac{p − 1}{p}\binom{p − 2}{t − 2}m^{p−t} (∑_{i=0}^{m−1} i)^{t−2} (m ∑_{i=0}^{m−1} i^{2} − (∑_{i=0}^{m−1} i)^{2})

= M\binom{p − 2}{t − 2} (\frac{m − 1}{2})^{t−2} ,

which is the coefficient of n^{p−t} in M (n + \frac{m−1}{2})^{p−2}. Therefore,

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (A_{p} − H_{p}) < \frac{1}{n^{p−1}} ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (\frac{A_{p} − H_{p}}{p} ∑_{k=1}^{p} \frac{∏_{j=1}^{p}(n + i_{j})}{n + i_{k}})

< \frac{M}{n^{p−1}} (n + \frac{m − 1}{2})^{p−2} .

Lemma 2. For every positive integer n ≥ M + \frac{(p−2)(m−1)}{2},

n^{p−1} ≥ M (n + \frac{m − 1}{2})^{p−2} .

Proof. For n ≥ M + \frac{(p−2)(m−1)}{2},

n^{p−1} ≥ Mn^{p−2} + \frac{(p − 2)(m − 1)}{2}n^{p−2}

≥ Mn^{p−2} + \frac{(p − 2)(m − 1)}{2}Mn^{p−3} + (\frac{(p − 2)(m − 1)}{2})^{2} n^{p−3}

⋮

≥ M ∑_{j=0}^{p−3} (\frac{(p − 2)(m − 1)}{2})^{j} n^{p−2−j} + (\frac{(p − 2)(m − 1)}{2})^{p−2} n

≥ M ∑_{j=0}^{p−2} (\frac{(p − 2)(m − 1)}{2})^{j} n^{p−2−j}.

Since (p − 2)^{j} ≥ \binom{p−2}{j} for all 0 ≤ j ≤ p − 2,

n^{p−1} ≥ M ∑_{j=0}^{p−2} \binom{p − 2}{j} (\frac{m − 1}{2})^{j} n^{p−2−j} = M (n + \frac{m − 1}{2})^{p−2}

as desired.

Now we are ready to state and prove the main result. Theorem 3. For every positive integer n ≥ M + \frac{(p−2)(m−1)}{2},

⌊\sqrt[p]{n} + \sqrt[p]{n + 1} + ⋯ + \sqrt[p]{n + m − 1}⌋ = ⌊\sqrt[p]{m^{p}n + \frac{m^{p}(m − 1)}{2} − 1}⌋ .

Proof. Let

S = \sqrt[p]{n} + \sqrt[p]{n + 1} + \sqrt[p]{n + 2} + ⋯ + \sqrt[p]{n + m − 1}.

One obtains

S^{p} = ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} \sqrt[p]{(n + i_{1}) ⋯ (n + i_{p})} = ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} G_{p}.

By the AM-GM-HM inequality,

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} A_{p} − ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1}(A_{p} − H_{p}) < S^{p} < ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} A_{p}.

Observe that

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} A_{p} = ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} n + ∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (\frac{1}{p} ∑_{j=1}^{p} i_{j})

= m^{p}n + \frac{m^{p}(m − 1)}{2}.

Moreover, by Lemma 1 and Lemma 2, we have

∑_{i_{1}=0}^{m−1} ⋯ ∑_{i_{p}=0}^{m−1} (A_{p} − H_{p}) < \frac{M}{n^{p−1}} (n + \frac{m − 1}{2})^{p−2} ≤ 1.

Thus

m^{p}n + \frac{m^{p}(m − 1)}{2} − 1 < S^{p} < m^{p}n + \frac{m^{p}(m − 1)}{2}

holds for all positive integers n ≥ M + \frac{(p−2)(m−1)}{2}. Equivalently,

⌊\sqrt[p]{n} + \sqrt[p]{n + 1} + ⋯ + \sqrt[p]{n + m − 1}⌋ = ⌊\sqrt[p]{m^{p}n + \frac{m^{p}(m − 1)}{2} − 1}⌋

holds for all positive integers n ≥ M + \frac{(p−2)(m−1)}{2}.

The table below compares the lower bounds of n between Saltzman and Yuan’s work, N_{SY} = \frac{m^{p}(m−1)(2m−1)(p−1)}{12p} + 1, and our work, N = \frac{(p−1)m^{p}(m^{2}−1)}{12p} + \frac{(p−2)(m−1)}{2}, for small values of p and m.

p = 3  p = 4  p = 5  p = 6
m N_{SY} N  N_{SY}  N  N_{SY}  N  N_{SY}  N
2 2  2  4  4  7  8  14  16
3 16 13  51  43  163  133  507  409
4 75 55  337  243  1434  1029  5974  4273
5 251 169 1407 942  7501  5006  39063  29050
6 661 423 4456 2840 28513 18152 178201 113410
7 1487 918 11705 7209 87397 53792 637266 392176
8 2587 1796 26881 16135 229377 137637 1911467 1146894
9 5509 3244 55769 32813 535378 314940 5019166 2952466
10 9501 5505 106876 61884 1140001 660014 11875001 6875018

3 Acknowledgments

The authors greatly appreciate the anonymous referee for his/her careful reading and helpful comments.

References

[1] F. D. Hammer, Problem E3010, Amer. Math. Monthly 95 (1988), 133–134.

[2] K. Kuhapatanakul and P. Ruankong, Sums of square roots, Math. Gazette 101 (2017), 276–280.

[3] P. W. Saltzman and P. Yuan, On sums of consecutive integral roots, Amer. Math. Monthly 115 (2008), 254–261.

[4] Z. Wang, A City of Nice Mathematics (in Chinese), The Democracy and Construction Press, Beijing, 2000.

[5] X. Zhan, Formulae for sums of consecutive square roots, Math. Intelligencer 27 (2005), 4–5.

2010 Mathematics Subject Classification : Primary 11A99; Secondary 11Y55. Keywords: sum of roots.

Received April 30 2019; revised version received September 2 2019. Published in Journal of Integer Sequences, September 25 2019.

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