SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the problem editor.

117. [1998, 117] Proposed by Kenneth B. Davenport, 301 Morea Road, Box 491, Frackville, Pennsylvania.

(a) Prove that

\begin{aligned}∫{∞ \atop 0} & \frac{1}{1 + x^{3}} · \frac{1}{1 + r^{3}x^{3}} ⋯ \frac{1}{1 + (r^{n}x)^{3}} dx \\ & = \frac{2π}{3\sqrt{3}} · \frac{(1 − r^{2})(1 − r^{5}) ⋯ (1 − r^{3n−1})}{(1 − r^{3})(1 − r^{6}) ⋯ (1 − r^{3n})}. \end{aligned}

(b) Prove that

\begin{aligned}∫{∞ \atop 0} & \frac{1}{1 + x^{4}} · \frac{1}{1 + r^{4}x^{4}} ⋯ \frac{1}{1 + (r^{n}x)^{4}} dx \\ & = \frac{π}{2\sqrt{2}} · \frac{(1 − r^{3})(1 − r^{7}) ⋯ (1 − r^{4n−1})}{(1 − r^{4})(1 − r^{8}) ⋯ (1 − r^{4n})}. \end{aligned}

Solution by Larry Eifler and Jerry Masuda, University of Missouri–Kansas City, Missouri.

Fix m > 1 and let r > 0. For n = 0, 1, 2, … and for x ≥ 0, define

\begin{aligned}ϕ_{n}(r, x) & = ∏_{k=0}^{n} \frac{1}{1 + (r^{k}x)^{m}} \\ & = \frac{1}{1 + x^{m}} · \frac{1}{1 + (rx)^{m}} ⋯ \frac{1}{(r^{n}x)^{m}} . \end{aligned}

Set

I_{n}(r) = ∫_{0}^{∞} ϕ_{n}(r, x)dx for n = 0, 1, 2, ….

It is well–known that (see Murray Spiegel, Complex Variables, McGraw–Hill, 1964, p. 185)

∫_{0}^{∞} \frac{y^{p}}{1 + y} \frac{dy}{y} = \frac{π}{sin(πp)} for 0 < p < 1

and so

I_{0}(r) = ∫_{0}^{∞} \frac{1}{1 + x^{m}} dx = \frac{1}{m} ∫_{0}^{∞} \frac{y^{1/m}}{1 + y} \frac{dy}{y} = \frac{π/m}{sin(π/m)}.

For n = 1, 2, …, we have

(1 + x^{m})ϕ_{n}(r, x) = ϕ_{n−1}(r, rx)

and

(1 + (r^{n}x)^{m})ϕ_{n}(r, x) = ϕ_{n−1}(r, x).

Thus,

I_{n−1}(r) = ∫_{0}^{∞} ϕ_{n−1}(r, x)dx = ∫_{0}^{∞} ϕ_{n−1}(r, rx)rdx

= ∫_{0}^{∞} r(1 + x^{m})ϕ_{n}(r, x)dx  (1)

and

I_{n−1}(r) = ∫_{0}^{∞}(1 + (r^{mn}x^{m}))ϕ_{n}(r, x)dx.  (2)

Using (1) and (2), we find

\begin{aligned}(1 − r^{mn−1})I_{n−1}(r) & = ∫_{0}^{∞}[(1 + (r^{mn}x^{m}) − r^{mn}(1 + x^{m})]ϕ_{n}(r, x)dx \\ & = (1 − r^{mn})I_{n}(r). \end{aligned}

It follows that

I_{n}(r) = \frac{π/m}{sin(π/m)} ∏_{k=1}^{n}(\frac{1 − r^{mk−1}}{1 − r^{mk}} ) for r ≠ 1.

Parts (a) and (b) of the problem are special cases of the above formula. Taking the limit as r approaches 1, we find that

∫_{0}^{∞}(\frac{1}{1 + x^{m}} )^{n}dx = \frac{π/m}{sin(π/m)} ∏_{k=1}^{n}(1 − \frac{1}{mk} ).

Also solved by the proposer.

118. [1998, 118] Proposed by Xuming Chen, 1126 Casson Green NW, Calgary, Alberta, Canada and Kenneth B. Davenport, 301 Morea Road, Box 491, Frackville, Pennsylvania.

Let

S(a, k, n) = ∑_{i=1}^{n} a^{i−1}i^{k}.

Show that

(1 − a)S(a, k, n) + ∑_{m=1}^{k} \binom{k}{m}(−1)^{m}S(a, k − m, n) = −a^{n}n^{k}.

Solution I by the proposers.

(1 − a)S(a, k, n) = ∑_{i=1}^{n} a^{i−1}i^{k} − ∑_{i=1}^{n} a^{i}i^{k}

= ∑_{i=1}^{n} a^{i−1}i^{k} − ∑_{j=2}^{n+1} a^{j−1}(j − 1)^{k}

= ∑_{i=1}^{n} a^{i−1}i^{k} − ∑_{j=1}^{n+1} a^{j−1}(j − 1)^{k}

= ∑_{i=1}^{n} a^{i−1}i^{k} − ∑_{j=1}^{n} a^{j−1}(j − 1)^{k} − a^{n}n^{k}

= ∑_{i=1}^{n} a^{i−1}i^{k} − ∑_{i=1}^{n} a^{i−1}(i − 1)^{k} − a^{n}n^{k}

= ∑_{i=1}^{n} a^{i−1}[i^{k} − (i − 1)^{k}]a^{i−1} − a^{n}n^{k}

= ∑_{i=1}^{n}[i^{k} − ∑_{m=0}^{k} \binom{k}{m}i^{k−m}(−1)^{m}]a^{i−1} − a^{n}n^{k}

= ∑_{i=1}^{n}[i^{k} − (i^{k} + ∑_{m=1}^{k} \binom{k}{m}i^{k−m}(−1)^{m})]a^{i−1} − a^{n}n^{k}

= ∑_{i=1}^{n} −( ∑_{m=1}^{k} \binom{k}{m}i^{k−m}(−1)^{m})a^{i−1} − a^{n}n^{k}

= − ∑_{m=1}^{k} \binom{k}{m}(−1)^{m}[∑_{i=1}^{n} i^{k−m}a^{i−1}] − a^{n}n^{k}

= − ∑_{m=1}^{k} \binom{k}{m}(−1)^{m}S(a, k − m, n) − a^{n}n^{k}.

Solution II by Jose Luis Diaz, Universitat Politecnica de Catalunya, Colom 1, Terrassa, Spain.

We observe that

P (a) = (1 − a)S(a, k, n) + ∑_{m=1}^{k} \binom{k}{m}(−1)^{m}S(a, k − m, n)

is a polynomial of degree n in the variable a. Hence,

P (a) = (1 − a)S(a, k, n) + \binom{k}{1}(−1)^{1}S(a, k − 1, n)

+ \binom{k}{2}(−1)^{2}S(a, k − 2, n) + ⋯ + \binom{k}{k}(−1)^{k}S(a, 0, n)

= \binom{k}{0}(−1)^{0}[1^{k} + 2^{k}a + ⋯ + n^{k}a^{n−1}]

+ \binom{k}{1}(−1)^{1}[1^{k−1} + 2^{k−1}a + ⋯ + n^{k−1}a^{n−1}]

+ \binom{k}{2}(−1)^{2}[1^{k−2} + 2^{k−2}a + ⋯ + n^{k−2}a^{n−1}] + ⋯

+ \binom{k}{k}(−1)^{k}[1^{0} + 2^{0}a + ⋯ + n^{0}a^{n−1}] − [1^{k}a + 2^{k}a^{2} + ⋯ + n^{k}a^{n}].

Now ordering the terms of P (a) in increasing powers of a we get the result. In fact,

P (a) = ∑_{j=0}^{k} \binom{k}{j}(−1)^{j} + [∑_{j=0}^{k} \binom{k}{j}(−1)^{j}2^{k−j} − 1^{k}]a

+ [∑_{j=0}^{k} \binom{k}{j}(−1)^{j}3^{k−j} − 2^{k}]a^{2} + ⋯ + [∑_{j=0}^{k} \binom{k}{j}(−1)^{j}n^{k−j} − n^{k}]a^{n−1} − n^{k}a^{n}

= (1 − 1)^{k} + [(2 − 1)^{k} − 1^{k}]a + [(3 − 1)^{k} − 2^{k}]a^{2} + ⋯ + [(n − 1)^{k} − (n − 1)^{k}]a^{n−1} − n^{k}a^{n} = −a^{n}n^{k}. Solution III by Carl Libis, 13126 Courbet Lane, Granada Hills, California. First note that

S(a, j + 1, n) = \frac{d}{da}[aS(a, j, n)] = S(a, j, n) + a\frac{d}{da}[S(a, j, n)]

for j = 0, 1, 2, . . .. Let S(a, ∗, n)^{j} := S(a, j, n). Now note that

(1 − a)S(a, k, n) + ∑_{m=1}^{k} \binom{k}{m}(−1)^{m}S(a, k − m, n) = −a^{n}n^{k}, k = 1, 2, 3, … (1)

is equivalent to

∑_{m=0}^{k} \binom{k}{m}(−1)^{m}S(a, k−m, n) = aS(a, k, n)−a^{n}n^{k} = aS(a, k, n−1), k = 1, 2, 3, …

which is equivalent to

(S(a, ∗, n) − 1)^{k} = aS(a, k, n − 1), k = 1, 2, 3, ….  (2)

To prove (1) we will prove (2) by induction. When k = 1, then

\begin{aligned}(S(a, ∗, n) − 1)^{1} & = S(a, 1, n) − S(a, 0, n) = ∑_{i=1}^{n} a^{i−1}(i − 1) = ∑_{i=2}^{n} a^{i−1}(i − 1) \\ & = ∑_{i=1}^{n−1} a^{i}i = a ∑_{i=1}^{n−1} a^{i−1}i = aS(a, 1, n − 1). \end{aligned}

Assume that (S(a, ∗, n) − 1)^{k} = aS(a, k, n − 1). Then

(S(a, ∗, n) − 1)^{k+1} = ∑_{m=0}^{k+1} \binom{k + 1}{m}(−1)^{m}S(a, k + 1 − m, n)

= ∑_{m=0}^{k+1}[\binom{k}{m − 1} + \binom{k}{m}](−1)^{m}S(a, k + 1 − m, n)

= ∑_{m=0}^{k+1} \binom{k}{m − 1}(−1)^{m}S(a, k + 1 − m, n) + ∑_{m=0}^{k+1} \binom{k}{m}(−1)^{m}S(a, k + 1 − m, n)

= ∑_{m=1}^{k+1} \binom{k}{m − 1}(−1)^{m}S(a, k + 1 − m, n) + ∑_{m=0}^{k} \binom{k}{m}(−1)^{m}S(a, k + 1 − m, n)

= ∑_{m=0}^{k} \binom{k}{m}(−1)^{m+1}S(a, k − m, n)

+ ∑_{m=0}^{k} \binom{k}{m}(−1)^{m}[S(a, k − m, n) + a\frac{d}{da}S(a, k − m, n)]

= a ∑_{m=0}^{k} \binom{k}{m}(−1)^{m} \frac{d}{da} ∑_{i=1}^{n} a^{i−1}i^{k−m}

= a ∑_{m=0}^{k} \binom{k}{m}(−1)^{m} ∑_{i=1}^{n}(i − 1)a^{i−2}i^{k−m}

= a ∑_{m=0}^{k} \binom{k}{m}(−1)^{m} ∑_{i=2}^{n}(i − 1)a^{i−2}i^{k−m}

= a ∑_{m=0}^{k} \binom{k}{m}(−1)^{m} ∑_{i=1}^{n−1} ia^{i−1}(i + 1)^{k−m}

= a ∑_{i=1}^{n−1} ia^{i−1} ∑_{m=0}^{k} \binom{k}{m}(−1)^{m}(i + 1)^{k−m}

= a ∑_{i=1}^{n−1} ia^{i−1}(i + 1 − 1)^{k}

= a ∑_{i=1}^{n−1} a^{i−1}i^{k+1}

= aS(a, k + 1, n − 1).

This completes the proof by induction of (2) and thus proves (1).

119. [1998, 118] Proposed by Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri.

If x and y are distinct positive real numbers, prove that

(\frac{\sqrt{x} + \sqrt{y}}{2} )^{2} < \frac{5(x + y)^{3}}{5(x + y)^{2} + (x − y)^{2}} < \frac{x − y}{sinh(\frac{x−y}{x+y} )}.

Solution I by the proposers.

On [0, 1], sinh θ < θ + θ^{3}/5. This can be easily proven by showing that f(θ) = sinh θ − θ − θ^{3}/5 is decreasing on [0, 1] and f(0) = 0. Without loss of generality, assume that x > y. So, since 0 < (x − y)/(x + y) < 1,

sinh(\frac{x − y}{x + y} ) < \frac{x − y}{x + y} + \frac{1}{5}(\frac{x − y}{x + y} )^{3}.

It is straightforward to show that this inequality is equivalent to

\frac{x − y}{sinh(\frac{x−y}{x+y} )} > \frac{5(x + y)^{3}}{5(x + y)^{2} + (x − y)^{2}} = \frac{5x^{3} + 15x^{2}y + 15xy^{2} + 5y^{3}}{6x^{2} + 6y^{2} + 8xy} .

The desired lower bound for the proposed inequality follows if it can be shown that

\frac{5x^{3} + 15x^{2}y + 15xy^{2} + 5y^{3}}{6x^{2} + 6y^{2} + 8xy} > \frac{(\sqrt{x} + \sqrt{y})^{2}}{2}

which is equivalent to showing that

x^{3} + y^{3} + 4x^{2}y + 4xy^{2} > \sqrt{xy}(3x^{2} + 3y^{2} + 4xy).

Squaring both sides of the latter inequality, simplifying and factoring gives (x − y)(x^{5} − y^{5}) > 0 – which is true since x > y and the desired inequalities follow.

Solution II by Joe Howard, New Mexico Highlands University, Las Vegas, New Mexico.

We use the method in Problem 567 of the College Mathematics Journal, Volume 28, Number 1 (1997), page 69. Let

z = \frac{x − y}{x + y} and \sqrt{1 − z^{2}} = \frac{2\sqrt{xy}}{x + y} .

Then the above inequality becomes

\frac{1 + \sqrt{1 − z^{2}}}{2} < \frac{5}{5 + z^{2}} < \frac{z}{sinh z}

or inverting

\frac{2(1 − \sqrt{1 − z^{2}})}{z^{2}} > 1 + \frac{z^{2}}{5} > \frac{sinh z}{z} = 1 + \frac{z^{2}}{3!} + \frac{z^{4}}{5!} + · · · .

Now

\frac{2(1 − \sqrt{1 − z^{2}})}{z^{2}} = 1 + \frac{z^{2}}{4} + \frac{1 · 3}{3!2}z^{4} + ⋯ + \frac{1 · 3 ⋯ (2n − 3)}{n!2^{n−1}}z^{2n−2} + ⋯

so the first inequality follows. For the 2nd inequality, we show

f(z) = z + \frac{z^{3}}{5} − sinh z > 0 for 0 < z < 1,

f(0) = 0,

f′(z) = 1 + \frac{3z^{2}}{5} − cosh z and f′(0) = 0

f′′(z) = \frac{6z}{5} − sinh z and f′′(0) = 0

f′′′(z) = \frac{6}{5} − cosh z and f′′′(0) = \frac{1}{5} > 0.

Hence, f(z) > 0 and the 2nd inequality follows.

Also solved by N. J. Kuenzi and Bob Prielipp, University of Wisconsin– Oshkosh, Oshkosh, Wisconsin; and Kandasamy Muthuvel, University of Wisconsin– Oshkosh, Oshkosh, Wisconsin.

120. [1998, 118] Proposed by Larry Hoehn, Austin Peay State University, Clarksville, Tennessee.

Let a, b, and c be the sides of a right triangle such that h is the altitude to hypotenuse c. Prove that c − 2h < a + b < 2c − h.

Solution I by Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; Joseph Wiener, University of Texas–Pan American, Edinburg, Texas; and Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

It is well known that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Hence, c < a + b so

c − 2h < a + b.  (∗)

By finding the area of the given triangle, we get (ab)/2 = (ch)/2. Since ab = ch and c^{2} = a^{2} + b^{2},

\begin{aligned}(2c − h)^{2} & = 4c^{2} − 4ch + h^{2} > 4(c^{2} − ch) \\ & = 4(a^{2} + b^{2} − ab) = (a + b)^{2} + 3(a − b)^{2} ≥ (a + b)^{2}. \end{aligned}

This implies that

2c − h > a + b.  (∗∗)

By (∗) and (∗∗), we get the required inequality

c − 2h < a + b < 2c − h.

Solution II by Hushang Poorkarimi and Donald Skow; University of Texas–Pan American, Edinburg, Texas.

Let c = x + y, where x is the projection of b to side c and y is the projection of a to side c. Then, b + h > x and a + h > y. Therefore, a + b + 2h > x + y so a + b + 2h > c. Hence,

a + b > c − 2h.  (1)

It is known that the altitude h is less than the median to side c, that is, h < c/2, and that a + b ≤ \sqrt{2}c which implies that a + b < (3c)/2. Thus, a + b < (3c)/2 and h < c/2 implies a + b + h < 2c. Hence,

a + b < 2c − h.  (2)

Inequalities (1) and (2) imply c − 2h < a + b < 2c − h.

Also solved by Joe Howard, New Mexico Highlands University, Las Vegas, New Mexico and the proposer.