Math Problems Vault

Problem 1.83

Let the functions

\varphi _0(t), \varphi _1(t), \varphi _2(t), \ldots , \varphi _n(t), \ldots

be non-negative in the interval $0<t<1$ and assume that

\varphi _0(t)+\varphi _1(t)+\varphi _2(t)+\cdots +\varphi _n(t)+\cdots = 1

holds for all $0<t<1$ . The sequence $s_0, s_1, s_2, \ldots , s_n, \ldots$ is supposed to be bounded. Construct the function

\Phi (t) = s_0 \varphi _0(t) + s_1 \varphi _1(t) + s_2 \varphi _2(t) + \cdots + s_n \varphi _n(t) + \cdots .

The range of $\Phi (t)$ will fall into the interval between the upper and lower bounds of the sequence $s_0, s_1, s_2, \ldots , s_n, \ldots$

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Problem 1.84

(continued). Show that

\lim _{t \rightarrow 1-0} \big( s_0 \varphi _0(t) + s_1 \varphi _1(t) + s_2 \varphi _2(t) + \cdots + s_n \varphi _n(t) + \cdots \big) = s

holds for every convergent sequence $s_0, s_1, s_2, \ldots$ for which $\lim_{n \rightarrow \infty } s_n = s$ if and only if for each fixed $\nu$

\lim _{t \rightarrow 1-0} \varphi _{\nu }(t) = 0 .

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Problem 1.85

The infinite sequences

a_0, a_1, a_2, \ldots , a_n, \ldots ; \qquad b_0, b_1, b_2, \ldots , b_n, \ldots

satisfy the following three conditions:

b_n>0, \quad n=0,1,2,\ldots ; \qquad b_0+b_1t+b_2t^2+\cdots \begin{cases} \text{is convergent for } \left|t\right|<1, \\ \text{divergent for } t=1 ; \end{cases}

\lim _{n \rightarrow \infty } \frac{a_n}{b_n} = s .

Then

a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots \text{ converges for } \left|t\right|<1

and

\lim _{t \rightarrow 1-0} \frac{a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots }{b_0+b_1t+b_2t^2+\cdots +b_nt^n+\cdots } = s .

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Remark.

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This proposition is due to E. Cesàro. Several applications will be given in the sequel.

Problem 1.86

If the series

a_0+a_1+a_2+\cdots +a_n+\cdots = s

is convergent then

\lim _{t \rightarrow 1-0} (a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots ) = s .

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Remark.

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N. H. Abel, l.c. solution 47, Vol. 1, p. 223.

Problem 1.87

Set

s_n = a_0+a_1+a_2+\cdots +a_n, \qquad n=0,1,2,3,\ldots

If

\lim _{n \rightarrow \infty } \frac{s_0+s_1+s_2+\cdots +s_n}{n+1} = s

exists, then

\lim _{t \rightarrow 1-0} (a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots ) = s .

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Remark.

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This proposition goes beyond problem 86 only if the series $a_0+a_1+a_2+\cdots +a_n+\cdots$ diverges (problem 67). G. Frobenius: J. reine angew. Math. Vol. 89, pp. 262-264 (1880).

Problem 1.88

If the following conditions are satisfied:

b_n>0, \qquad \sum _{n=0}^{\infty } b_n \text{ divergent}, \qquad \lim _{n \rightarrow \infty } \frac{a_0+a_1+a_2+\cdots +a_n}{b_0+b_1+b_2+\cdots +b_n} = s ,

then

\lim _{t \rightarrow 1-0} \frac{a_0+a_1t+a_2t^2+\cdots +a_nt^n+\cdots }{b_0+b_1t+b_2t^2+\cdots +b_nt^n+\cdots } = s ,

provided that the series in the denominator converges for $\left|t\right|<1$ .

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Problem 1.89

The following limit exists and is positive provided $\alpha$ is positive:

\lim _{t \rightarrow 1-0} (1-t)^{\alpha } \big( 1^{\alpha -1}t + 2^{\alpha -1}t^2 + 3^{\alpha -1}t^3 + \cdots + n^{\alpha -1}t^n + \cdots \big) .

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Problem 1.90

If $0<k<1$ and if $k$ converges to $1$ , then

\int _0^1 \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} \sim \frac{1}{2} \log \frac{1}{1-k} . \qquad \text{[See Problem 202 of Part II.]}

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Problem 1.91

Let $A_n$ and $B_n$ be the numerator and denominator, respectively, of the $n$ -th convergent of the infinite continued fraction

\cfrac {a}{1} + \cfrac {a}{3} + \cfrac {a}{5} + \cdots \quad \text{and so} \quad \frac{A_n}{B_n} = \cfrac {a}{1} + \cfrac {a}{3} + \cfrac {a}{5} + \cdots + \cfrac {a}{2n-3} , \quad a>0 .

Assume that this continued fraction converges. Find its value applying problem 85 and using the series

F(x) = \sum _{n=0}^{\infty } \frac{A_n}{n!} x^n , \qquad G(x) = \sum _{n=0}^{\infty } \frac{B_n}{n!} x^n .

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Remark.

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$F(x)$ and $G(x)$ satisfy a linear homogeneous differential equation of the second order by virtue of the recursion formulas for $A_n$ and $B_n$ . Cf.O. Perron: Die Lehre von den Kettenbrüchen. Leipzig: B. G. Teubner 1913, p. 353, formula (24); R. G. Archibald: An Introduction to the Theory of Numbers. Columbus/Ohio: Charles E. Merrill Publishing 1970, p. 176.