The Structure of Real Sequences and Series

Let the terms of the sequence $a_1, a_2, a_3, \ldots$ satisfy the condition

then the sequence

either converges to its lower bound or diverges properly to $-\infty$.

Assume that the terms of the sequence $a_1, a_2, a_3, \ldots$ satisfy the condition

Then

exists; $\omega$ is finite and we have

If the general term of a series which is neither convergent nor properly divergent tends to $0$ the partial sums are everywhere dense between their lowest and their highest limit points.

Let $a_n>0$, $\lim_{n \rightarrow \infty } a_n = 0$ and the series $a_1+a_2+\cdots +a_n+\cdots$ be divergent. Put $a_1+a_2+\cdots +a_n = s_n$ and denote by $\left[s_n\right]$ the largest integer $\leq s_n$. Find the limit points of the sequence

Assume that there exists for the sequence $t_1, t_2, \ldots , t_n, \ldots$ a sequence of positive numbers $\varepsilon_1, \varepsilon_2, \ldots , \varepsilon_n, \ldots$, converging to $0$, for which

Then the numbers $t_1, t_2, \ldots , t_n, \ldots$ are everywhere dense between their lowest and highest limit points.

Let $\nu_1, \nu_2, \ldots , \nu_n, \ldots$ be positive integers, $\nu_1 \leq \nu_2 \leq \nu_3 \leq \cdots$. The set of limit points of the sequence

consists of a closed interval (of length 0 if the limit exists).

A subsequence whose terms are the successive partial sums of an absolutely convergent series can be picked out from every convergent sequence.

A sequence $t_1, t_2, \ldots , t_n, \ldots$ that diverges to $+\infty$ contains a minimum (i.e.there exists a $t_n$ such that $t_m \geq t_n$ for all $m$).

A convergent sequence has either a maximum or a minimum or both.

Let $l_1, l_2, l_3, \ldots , l_m, \ldots$ be a sequence of positive numbers (positive in the sense of $>0$) and let $\liminf_{m \rightarrow \infty } l_m = 0$. Then there are infinitely many subscripts $n$ for which $l_n$ is smaller than all the terms $l_1, l_2, l_3, \ldots , l_{n-1}$ preceding $l_n$, ($l_n < l_k$, $k=1,2,3,\ldots ,n-1$).

Let $l_1, l_2, l_3, \ldots , l_m, \ldots$ be a sequence of positive numbers (positive in the sense of $>0$) and let $\lim_{m \rightarrow \infty } l_m = 0$. Then there are infinitely many subscripts $n$ for which $l_n$ is larger than all the terms $l_{n+1}, l_{n+2}, \ldots$ following $l_n$, ($l_n > l_{n+k}$, $k=1,2,3,\ldots$). (Not only the conclusion but also the hypothesis is different from the one in problem 107.)

Given two sequences

satisfying the conditions

Then there are infinitely many subscripts $n$ such that two different kinds of inequalities hold simultaneously:

If the sequence $\dfrac {L_1}{1}, \dfrac {L_2}{2}, \ldots , \dfrac {L_m}{m}, \ldots$ tends to $+\infty$ and if $A$ is larger than its minimum [problem 105] then there exists a subscript $n$ (or several subscripts $n$), $n \geq 1$, so that the quotients

are $\leq A$ and the infinitely many quotients

are all $\geq A$.

Assume that the sequence $l_1, l_2, l_3, \ldots , l_m, \ldots$ satisfies the sole condition

Let $A$ be larger than $l_1$ ($A>l_1$). Then there exists a subscript $n$, $n \geq 1$, such that all these inequalities hold simultaneously:

If $A$ tends to infinity then so does $n$.

Let the sequence $l_1, l_2, \ldots , l_m, \ldots$ satisfy the two conditions

and assume $l_1>A>0$. Then there exists a subscript $n$, $n \geq 1$, such that the inequalities

hold simultaneously. If $A$ tends to $0$, then $n$ tends to infinity.

Show that

Let $x_1, x_2, x_3, \ldots , x_m, \ldots$ be arbitrary real numbers, $x_m \neq 0$. If there exists a positive distance $\delta$ such that $\left|x_l-x_k\right|>\delta$, $l<k$, $l,k=1,2,3,\ldots$, then the convergence exponent of $\left|x_1\right|, \left|x_2\right|, \left|x_3\right|, \ldots , \left|x_m\right|, \ldots$ is at most $1$.

Let $\beta$ be larger than the convergence exponent of the sequence $r_1, r_2, \ldots$. Then there exist infinitely many subscripts $n$ for which the $n-1$ inequalities

are satisfied.

Assume that the convergence exponent $\lambda$ of the sequence $r_1, r_2, r_3, \ldots , r_m, \ldots$ is positive and $0<\alpha <\lambda <\beta$. Then there are infinitely many subscripts such that the two types of inequalities

hold simultaneously.

Suppose $0<r_1<r_2<r_3<\cdots$. For what values of $x$, $x \geq 0$, is the $m$-th term of the series

larger than all the other terms, $m=0,1,2,\ldots$?

Assume

Then arbitrarily large values of $n$ and $r$ can be found for which the following inequalities hold simultaneously [problem 111]:

For an everywhere convergent power series in $x$ which is not merely a polynomial the central subscript $\nu (x)$ tends to $\infty$ with $x$.

The subscript of the maximum term increases as $x$ increases.

The series

with positive coefficients and finite radius of convergence $\varrho$ ($p_m>0$, $\varrho >0$) is such that one term after the other, all terms in turn, become maximum term. Then $\dfrac {1}{\varrho }$ is the radius of convergence of the series

The dominance of the maximum term is more pronounced in an always convergent series than in one that does not always converge (it is strongest in a polynomial). More exactly: let the radius of convergence of the power series

be infinite and that of the power series

be finite. Suppose $a_m \geq 0$, $b_m>0$, $m=0,1,2,\ldots$ The coefficients $b_0, b_1, b_2, \ldots$ be such that all the terms $b_m y^m$ become in turn maximum term [problem 120]. Then a value $\bar{y}$ can be determined for certain arbitrarily large positive $\bar{x}$ such that for these corresponding values the respective series have the same central subscript. Let the common central subscript be $n$. Then all the following inequalities hold simultaneously:

If there are values $x^*$ to which no $y$ corresponds in the sense of problem 122 then they are ``rare''. They have a finite logarithmic measure, i.e.the set of points $\log x^*$, $x^*$ exceptional value, may be covered by countably many intervals of finite total length.

From the harmonic series

remove all terms that contain the digit 9 in the decimal representation of the denominator. The resulting subseries is convergent.

If all the subseries of a series converge then the series is absolutely convergent.

Let $k$ and $l$ denote positive integers. Must the convergent series $a_1+a_2+a_3+\cdots$ be absolutely convergent if all its subseries of the form

(subscripts in arithmetic progression) converge?

Let $k$ and $l$ be integers, $k \geq 1$, $l \geq 2$. Must the convergent series $a_1+a_2+a_3+\cdots$ be absolutely convergent if all its subseries of the form

(subscripts in geometric progression) are convergent?

Let $\varphi (x)$ denote a polynomial assuming integral values for integral $x$, $\varphi (x) = c_0 x^l + c_1 x^{l-1} + \cdots$ Assume that the degree is $l \geq 1$ and that the coefficient $c_0$ of $x^l$ is positive ($c_0>0$). The values $\varphi (0), \varphi (1), \varphi (2), \ldots$ form a generalized arithmetic progression of order $l$; since $c_0>0$ only a finite number of terms of the progression can be negative. Must a convergent series $a_1+a_2+a_3+\cdots$ converge absolutely if all its subseries whose subscripts form a generalized arithmetic progression

(omitting the terms with negative subscripts) converge?

If the series $a_1+a_2+a_3+\cdots$ converges absolutely and if every subseries

has the sum $0$ then $a_1=a_2=\cdots =0$.

Consider the set of points determined by all the subseries of

This set is perfect and nowhere dense (closed and dense in itself, but nowhere dense in the set of all real numbers).

Let the terms of the convergent series

satisfy the inequalities

Then it is possible to represent any number $\sigma$ in the half-closed interval $0<\sigma \leq s$ by an infinite subseries:

Find the series $p_1+p_2+p_3+\cdots$ that satisfies the conditions

and verify that in this case every $\sigma$ mentioned in problem 131 can be represented by one infinite subseries only.

By rearranging the factors of the infinite product

we obtain the infinite product

in which blocks of $p$ factors greater than 1 alternate with blocks of $q$ factors smaller than 1. (Factors of the same kind remain in the ``natural'' order.) Show that

By rearranging the factors of the infinite product

we obtain the infinite product

in which blocks of $p$ factors greater than 1 alternate with blocks of $q$ factors smaller than 1. (Factors of the same kind remain in the ``natural'' order.) Show that

By rearranging the terms of the infinite series

we obtain the infinite series

in which blocks of $p$ positive terms alternate with blocks of $q$ negative terms. (Terms of the same kind remain in the ``natural'' order, steadily decreasing in absolute value.) Show that

By rearranging the terms of the infinite series

we obtain the infinite series

in which blocks of $p$ positive terms alternate with blocks of $q$ negative terms. (Terms of the same kind remain in the ``natural'' order, steadily decreasing in absolute value.) Show that

If one of the two to each other complementary subseries of a convergent series is convergent the other is convergent too. A rearrangement which only shifts the two subseries relatively to each other does not change the sum of such a series.

If one of the two to each other complementary subseries of a conditionally convergent series diverges to $+\infty$ then the other diverges to $-\infty$. Provided that all the terms of one of these two subseries are of the same sign it is possible to obtain an arbitrary sum for the whole series by shifting the two subseries relatively to each other.

It is not possible to accelerate by rearrangement the divergence of a divergent series with positive monotone decreasing terms.

By rearranging the series we can slow down arbitrarily the divergence of a divergent series with positive terms which tend to $0$. More explicitly: assume

Then there exists a rearranged series $p_{\nu_1}+p_{\nu_2}+p_{\nu_3}+\cdots +p_{\nu_n}+\cdots$ such that $p_{\nu_1}+p_{\nu_2}+\cdots +p_{\nu_n} \leq Q_n$ for $n=1,2,3,\ldots$

Assume that

and $\left|a_1\right|+\left|a_2\right|+\left|a_3\right|+\cdots +\left|a_n\right|+\cdots$ is divergent. Let $s'<s<s''$. By a rearrangement that leaves all the negative terms at their places the sum $s'$ can be realized; by a rearrangement that leaves all the positive terms at their places the sum $s''$ can be realized [problem 136].

Assume that $p_n>0$, $p_1 \geq p_2 \geq p_3 \geq \cdots$ and that the series

is divergent and the series

where $\varepsilon_k$ is $-1$ or $+1$, is convergent. If under these conditions a certain percentage of terms is positive then it is $50\%$. More precisely:

Suppose $p_n>0$, $p_1 \geq p_2 \geq p_3 \geq \cdots$ and that the series

where $\varepsilon_k$ is $-1$ or $+1$, is convergent. Then