SOLUTIONS

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105. [1997, 105] Proposed by Kenneth Davenport, P. O. Box 99901, Pittsburgh, Pennsylvania.

Evaluate the series

\frac{3}{1} + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + ⋯

where the denominators are the triangular numbers and every two terms the signs alternate, i.e. +, +, −, −, +, +, etc.

Solution I by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri; Joe Howard, New Mexico Highlands University, Las Vegas, New Mexico; Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin; and the proposer.

The triangular numbers, {T_{n}}_{n=1}^{∞}, have the form T_{n} = n(n+1)/2. The partial sum, S_{2n}, of the first 2n terms of the series is

S_{2n} = 2 + ∑_{k=0}^{n−1}[\frac{2}{(2k + 1)(2k + 2)} + \frac{2}{(2k + 2)(2k + 3)}](−1)^{k}.

In this replace

\frac{1}{(2k + 1)(2k + 2)} by \frac{1}{2k + 1} − \frac{1}{2k + 2}

and

\frac{1}{(2k + 2)(2k + 3)} by \frac{1}{2k + 2} − \frac{1}{2k + 3}.

We obtain

S_{2n} = 2 + 2 ∑_{k=0}^{n−1}[\frac{1}{2k + 1} − \frac{1}{2k + 3}](−1)^{k}.

If the summation is written out, all terms are doubled except those for k = 0 and k = n − 1. This gives

\begin{aligned}S_{2n} & = 2 + 2[2 ∑_{k=0}^{n−1} \frac{1}{2k + 1}(−1)^{k} + \frac{(−1)^{n}}{2n + 1} − 1] \\ & = 4 ∑_{k=0}^{n−1} \frac{1}{2k + 1}(−1)^{k} + \frac{2(−1)^{n}}{2n + 1} . \end{aligned}

The indicated summation is a partial sum of the Leibniz–Gregory series for tan^{−1} 1. Hence,

lim_{n→∞} S_{2n} = 4 ∑_{k=0}^{∞} \frac{1}{2k + 1}(−1)^{k} = 4(\frac{π}{4} ) = π.

Solution II by Carl Libis, University of Alabama, Tuscaloosa, Alabama.

\frac{3}{1} + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + ⋯ = 2[\frac{3}{1 · 2} + \frac{1}{2 · 3} − \frac{1}{3 · 4} − \frac{1}{4 · 5} + ⋯]

= 2[(\frac{3}{1} − \frac{3}{2}) + (\frac{1}{2} − \frac{1}{3}) − (\frac{1}{3} − \frac{1}{4}) − (\frac{1}{4} − \frac{1}{5}) + ⋯]

= 2[3 − 1 − \frac{2}{3} + \frac{2}{5} − \frac{2}{7} + ⋯]

= 4[\frac{1}{1} − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + ⋯]

= 4[∑_{n=0}^{∞} \frac{1}{4n + 1} − ∑_{n=0}^{∞} \frac{1}{4n + 3}].

Let z = 1/4 in the formula

π cot πz = \frac{1}{z} + ∑_{n=1}^{∞} \frac{2z}{z^{2} − n^{2}} .

We obtain

π = 4 + \frac{1}{2} ∑_{n=1}^{∞} \frac{1}{(1/4)^{2} − n^{2}}

= 4 + 8 ∑_{n=1}^{∞} \frac{1}{1 − 16n^{2}}

= 4 + 4 ∑_{n=1}^{∞}(\frac{1}{1 + 4n} + \frac{1}{1 − 4n})

= 4(1 + ∑_{n=1}^{∞} \frac{1}{4n + 1} − ∑_{n=1}^{∞} \frac{1}{4n − 1})

= 4(∑_{n=0}^{∞} \frac{1}{4n + 1} − ∑_{n=0}^{∞} \frac{1}{4n + 3}).

Therefore,

\frac{3}{1} + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + ⋯ = π.

Solution III by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

We begin by establishing the following preliminary result.

\underline{Lemma}.

1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + ⋯ + \frac{1}{n(n + 1)/2} + \frac{1}{(n + 1)(n + 2)/2} + ⋯ = 2.

\underline{Proof}.

1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + ⋯ + \frac{1}{n(n + 1)/2} + \frac{1}{(n + 1)(n + 2)/2} + ⋯

= lim_{n→∞}(1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + ⋯ + \frac{1}{n(n + 1)/2} + \frac{1}{(n + 1)(n + 2)/2})

= lim_{n→∞}[(\frac{2}{1} − \frac{2}{2}) + (\frac{2}{2} − \frac{2}{3}) + (\frac{2}{3} − \frac{2}{4}) + (\frac{2}{4} − \frac{2}{5}) + ⋯

+ (\frac{2}{n} − \frac{2}{n + 1}) + (\frac{2}{n + 1} − \frac{2}{n + 2})]

= lim_{n→∞}(2 − \frac{2}{n + 2}) = 2.

\underline{Corollary}.

1 + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + \frac{1}{15} + \frac{1}{21} − \frac{1}{28} − \frac{1}{36} + ⋯

converges absolutely.

Recalling that

1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + \frac{1}{9} − ⋯ = \frac{π}{4} ,

it follows that

(\frac{1}{3} − \frac{1}{5}) + (\frac{1}{7} − \frac{1}{9}) + ⋯ + (\frac{1}{4n − 1} − \frac{1}{4n + 1}) + ⋯ = 1 − \frac{π}{4} .

Thus,

1 + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + \frac{1}{15} + \frac{1}{21} − \frac{1}{28} − \frac{1}{36} + ⋯

= 1 + \frac{1}{3} + (\frac{1}{6} − \frac{1}{3}) + (\frac{1}{10} − \frac{1}{5}) + \frac{1}{15} + \frac{1}{21} + (\frac{1}{28} − \frac{1}{14}) + (\frac{1}{36} − \frac{1}{18}) + ⋯

= [1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \frac{1}{28} + \frac{1}{36} + ⋯]

− [(\frac{1}{3} + \frac{1}{5}) + (\frac{1}{14} + \frac{1}{18}) + ⋯]

= 2 − 4(\frac{2}{3 · 5} + \frac{2}{7 · 9} + ⋯ + \frac{2}{(4n − 1)(4n + 1)} + ⋯)

= 2 − 4[(\frac{1}{3} − \frac{1}{5}) + (\frac{1}{7} − \frac{1}{9}) + ⋯ + (\frac{1}{4n − 1} − \frac{1}{4n + 1}) + ⋯]

= 2 − 4(1 − \frac{π}{4} ) = −2 + π.

Therefore,

\frac{3}{1} + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + ⋯

= 2 + (1 + \frac{1}{3} − \frac{1}{6} − \frac{1}{10} + ⋯)

= 2 + (−2 + π) = π.

106. [1997, 105] Proposed by Mohammad K. Azarian, University of Evansville, Evansville, Indiana.

Show that

\frac{1}{2} ∑_{m=0}^{n} \binom{n}{m}\frac{1}{(n − m + 1)(m + 1)} = \frac{2^{n+1} − 1}{(n + 2)(n + 1)}.

Solution I by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri and Carl Libis, University of Alabama, Tuscaloosa, Alabama.

\binom{n}{m} = \frac{n(n − 1) ⋯ (n − m + 1)}{m!},

so

\binom{n}{m}\frac{1}{(n − m + 1)(m + 1)} = \frac{1}{(n + 1)(n + 2)}\binom{n + 2}{m + 1}.

Hence,

\frac{1}{2} ∑_{m=0}^{n} \binom{n}{m}\frac{1}{(n − m + 1)(m + 1)} = \frac{1/2}{(n + 1)(n + 2)} ∑_{m=0}^{n} \binom{n + 2}{m + 1}

= \frac{1/2}{(n + 1)(n + 2)} ∑_{m=1}^{n+1} \binom{n + 2}{m}

= \frac{1/2}{(n + 1)(n + 2)}[ ∑_{m=0}^{n+2} \binom{n + 2}{m} − \binom{n + 2}{n + 2} − \binom{n + 2}{0}]

= \frac{1/2}{(n + 1)(n + 2)}(2^{n+2} − 2)

= \frac{2^{n+1} − 1}{(n + 1)(n + 2)}.

Solution II by Bob Prielipp, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

We begin by establishing the following preliminary result. \underline{Lemma}.

\frac{1}{n + 1}\binom{n + 1}{m + 1} = \frac{1}{m + 1}\binom{n}{m}.

\underline{Proof}.

\begin{aligned}\frac{1}{n + 1}\binom{n + 1}{m + 1} & = \frac{1}{n + 1} · \frac{(n + 1)!}{(m + 1)!((n + 1) − (m + 1))!} \\ & = \frac{1}{m + 1} · \frac{n!}{m!(n − m)!} = \frac{1}{m + 1}\binom{n}{m}. \end{aligned}

\underline{Corollary}.

∑_{m=0}^{n} \frac{1}{m + 1}\binom{n}{m} = \frac{1}{n + 1}(2^{n+1} − 1).

\underline{Proof}.

\begin{aligned}∑_{m=0}^{n} \frac{1}{m + 1}\binom{n}{m} & = ∑_{m=0}^{n} \frac{1}{n + 1}\binom{n + 1}{m + 1} = \frac{1}{n + 1} ∑_{m=0}^{n} \binom{n + 1}{m + 1} \\ & = \frac{1}{n + 1}( ∑_{m=0}^{n+1} \binom{n + 1}{m} − \binom{n + 1}{0}) = \frac{1}{n + 1}(2^{n+1} − 1). \end{aligned}

Thus,

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

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

= \frac{1}{2} · \frac{1}{n + 2}( ∑_{m=0}^{n} \frac{1}{m + 1}\binom{n}{m} + ∑_{m=0}^{n} \frac{1}{n − m + 1}\binom{n}{n − m})

= \frac{1}{2} · \frac{1}{n + 2}(2 ∑_{m=0}^{n} \frac{1}{m + 1}\binom{n}{m})

= \frac{1}{n + 2} · \frac{1}{n + 1}(2^{n+1} − 1) = \frac{2^{n+1} − 1}{(n + 1)(n + 2)}.

Solution III by Russell Euler and Jawad Sadek, Northwest Missouri State University, Maryville, Missouri and Kenneth Davenport, P. O. Box 99901, Pittsburg, Pennsylvania.

First notice that

∑_{m=0}^{n} \binom{n}{m}\frac{1}{(n − m + 1)(m + 1)} = \frac{1}{n + 2} ∑_{m=0}^{n} \binom{n}{m}[\frac{1}{n − m + 1} + \frac{1}{m + 1}]. (1)

Integrating the identity

∑_{m=0}^{n} \binom{n}{m}x^{m} = (1 + x)^{n}

from 0 to 1 gives

∑_{m=0}^{n} \binom{n}{m}\frac{1}{m + 1} = \frac{2^{n+1} − 1}{n + 1}.  (2)

Similarly, since

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

∑_{m=0}^{n} \binom{n}{m}\frac{1}{n − m + 1} = \frac{2^{n+1} − 1}{n + 1}.  (3)

Substituting (2) and (3) into the right hand side of (1) and simplifying yields

\frac{1}{2} ∑_{m=0}^{n} \binom{n}{m}\frac{1}{(n − m + 1)(m + 1)} = \frac{2^{n+1} − 1}{(n + 2)(n + 1)}.

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

The Binomial Series is

∑_{m=0}^{n} \binom{n}{m}x^{m}y^{n−m} = (x + y)^{n}.

Integrating with respect to dx and dy gives

∑_{m=0}^{n} \binom{n}{m} ∫_{0}^{1} x^{m}dx ∫_{0}^{1} y^{n−m}dy = ∫_{0}^{1} ∫_{0}^{1}(x + y)^{n}dxdy.

Therefore,

∑_{m=0}^{n} \binom{n}{m}\frac{1}{(m + 1)(n − m + 1)} = \frac{1}{n + 1}[\frac{(1 + y)^{n+2} − y^{n+2}}{n + 2} ]_{0}^{1} = \frac{2^{n+2} − 2}{(n + 2)(n + 1)}.

Solution V by the proposer.

First we note that if

f(x) = a_{0} + a_{1} \frac{x}{1!} + a_{2} \frac{x^{2}}{2!} + a_{3} \frac{x^{3}}{3!} + ⋯ + a_{n} \frac{x^{n}}{n!} + ⋯

and

g(x) = b_{0} + b_{1} \frac{x}{1!} + b_{2} \frac{x^{2}}{2!} + b_{3} \frac{x^{3}}{3!} + ⋯ + b_{n} \frac{x^{n}}{n!} + ⋯,

then the coefficient of x^{n}/n! in the product f(x)g(x) is

∑_{m=0}^{n} \binom{n}{m}a_{n−m}b_{m}.

Now, if

f(x) = g(x) = \frac{e^{x} − 1}{x} = 1 + \frac{x}{2!} + \frac{x^{2}}{3!} + \frac{x^{3}}{4!} + ⋯ + \frac{x^{n−1}}{n!} + ⋯,

then by the above note the coefficient of x^{n}/n! in

f(x)g(x) = \frac{(e^{x} − 1)^{2}}{x^{2}}

is

∑_{m=0}^{n} \binom{n}{m}\frac{1}{n − m + 1} · \frac{1}{m + 1}.  (1)

On the other hand,

f(x)g(x) = \frac{(e^{x} − 1)^{2}}{x^{2}} = \frac{1}{x^{2}} (e^{2x} − 2e^{x} + 1)

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

Thus, the coefficient of x^{n}/n! in f(x)g(x) using equation (2) is

\frac{2^{n+2} − 2}{(n + 2)(n + 1)}.  (3)

Consequently, from (1) and (3) we obtain the desired equality.

107. [1997, 106] Proposed by Leonard L. Palmer, Southeast Missouri State University, Cape Girardeau, Missouri.

Prove, if p = 8k ± 3 is a prime for k ≥ 1 and

a^{2} + (p − 2)b^{2} ≡ 0 (mod p),

then a ≡ 0 (mod p) and b ≡ 0 (mod p).

Solution I by James T. Bruening, Southeast Missouri State University, Cape Girardeau, Missouri and Kandasamy Muthuvel, University of Wisconsin–Oshkosh, Oshkosh, Wisconsin.

The congruence,

a^{2} + (p − 2)b^{2} ≡ 0 (mod p),

can be rewritten as

a^{2} + pb^{2} − 2b^{2} ≡ 0 (mod p),

or,

a^{2} ≡ 2b^{2} (mod p).

Assume p |̸b, and let r be a quadratic residue modulo p such that b^{2} ≡ r (mod p). Then (r/p) = 1, where (x/p) denotes the Legendre symbol. a^{2} ≡ 2b^{2} (mod p) implies a^{2} ≡ 2r (mod p), so that (2r/p) = 1. But

(2r/p) = (2/p)(r/p) = (−1) · (1) = −1,

since p ≡ ±3 (mod 8) implies (2/p) = −1. (See [1], Section 9.2, pp. 180–187.)

Since (2r/p) cannot equal both 1 and -1, this is a contradiction, so the assumption, p |̸b, is false. Thus p|b, and b ≡ 0 (mod p). a^{2} ≡ 2b^{2} (mod p) implies a^{2} ≡ 0 (mod p), so that a ≡ 0 (mod p), since p is prime. This completes the proof.

\underline{Reference}

1. D. M. Burton, Elementary Number Theory, 3rd ed., McGraw–Hill, New York, 1997.

Solution II by the proposer. Let p = 8k + 3. Because 1 = (8k + 1)(4k + 1) + (8k + 3)(−4k) we have (4k + 1)(8k + 1) ≡ 1 (mod p). Using Legendre symbols and (4k + 1)^{2} = (2k + 1) + 2k(8k + 3) we have

(\frac{2k + 1}{8k + 3}) = 1

and

(\frac{4k + 2}{8k + 3}) = (\frac{2}{8k + 3})(\frac{2k + 1}{8k + 3}) = (−1)(1) = −1.

Suppose a^{2} + (p − 2)b^{2} ≡ 0 (mod p) and a ≢ 0 (mod p). There exists a c such that ac ≡ 1 (mod p) so (ac)^{2} + (p − 2)(bc)^{2} ≡ 0 (mod p). Therefore, (8k + 1)(bc)^{2} ≡ −1 (mod p) and (4k+1)(8k+1)(bc)^{2} ≡ −(4k+1) ≡ 4k+2 (mod p). But (bc)^{2} ≡ 4k+2 (mod p) is a contradiction because

(\frac{4k + 2}{8k + 3}) = −1.

So a ≡ 0 (mod p) and b^{2} ≡ 0 (mod p), which implies b ≡ 0 (mod p).

Similarly, let p = 8k + 3. We have 1 = (8k − 5)(4k − 2) + (8k − 3)(−4k + 3) implying (4k − 2)(8k − 5) ≡ 1 (mod p). Also (4k − 1)^{2} = (−2k + 1) + 2k(8k − 3) which gives

(\frac{−2k + 1}{8k − 3}) = 1

and

(\frac{−4k + 2}{8k − 3}) = (\frac{2}{8k − 3})(\frac{−2k + 1}{8k − 3}) = (−1)(1) = −1.

As in the preceding case there is a c such that (p − 2)(bc)^{2} ≡ −1 (mod p) or (8k − 5)(bc)^{2} ≡ −1 (mod (8k − 3)). We have (4k − 2)(8k − 5)(bc)^{2} ≡ −(4k − 2) ≡ −4k +2 (mod p) so (bc)^{2} ≡ −4k +2 (mod p). Again this is a contradiction so a ≡ 0 (mod p) and b ≡ 0 (mod p).

108. [1997, 106] Proposed by Joseph B. Dence, University of Missouri–St. Louis, St. Louis, Missouri.

A positive integer d is called a unitary divisor of a positive integer n, written d || n, if d and n/d are relatively prime. We define two unitary arithmetic functions by analogy to their standard counterparts:

A unitary Möbius function μ^{∗}(n):

∑_{d || n} μ^{∗}(d) = \{ \begin{matrix}1, & for n = 1; \\ 0, & for n > 1. \end{matrix}

A unitary Euler phi-function ϕ^{∗}(n):

ϕ^{∗}(n) = ∑_{d || n} μ^{∗}(d)\frac{n}{d}.

When n > 2, ϕ(n) is always even; this is not true of ϕ^{∗}(n). Determine how many known odd primes are in the range of the function ϕ^{∗}(n).

Solution by the proposer. It is straightforward to show that μ^{∗}(n) is multiplicative: n = m_{1}m_{2}, (m_{1}, m_{2}) = 1 implies μ^{∗}(n) = μ^{∗}(m_{1})μ^{∗}(m_{2}), and from this fact that ϕ^{∗}(n) is also multiplicative. Next we note that if p is an odd prime and k ≥ 1, then

ϕ^{∗}(p^{k}) = μ^{∗}(1)\frac{p^{k}}{1} + μ^{∗}(p^{k})\frac{p^{k}}{p^{k}} = p^{k} − 1,

so if n contains one or more odd primes in its factorization, ϕ^{∗}(n) is even. Hence, ϕ^{∗}(n) can only be an odd prime if n = 2^{k}, in which case ϕ^{∗}(2^{k}) = 2^{k} − 1. Primes of this form are the Mersenne primes (e.g., 3, 7, 31, 127, …); at present, there are approximately three dozen known Mersenne primes.