This section owes its extraordinary appearance to many, many authors, reading whose works I wanted to launch these works into the writers themselves. Actually, I planned to publish this topic in full only when it was finally ready, but due to too large quantity questions about it, I will outline some points now. Subsequently, the material will be supplemented and expanded. Let's start with definitions.
A series of the form $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$, where $u_n>0$, is called alternating.
The signs of the members of the alternating series strictly alternate:
$$ \sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n=u_1-u_2+u_3-u_4+u_5-u_6+u_7-u_8+\ldots $$
For example, $1-\frac(1)(2)+\frac(1)(3)-\frac(1)(4)+\ldots$ is an alternating series. It happens that a strict alternation of signs does not begin with the first element, but this is not significant for convergence studies.
Why is alternating characters not starting with the first element unimportant? show\hide
The fact is that among the properties of number series there is a statement that allows us to discard “extra” members of the series. This is the property:
The series $\sum\limits_(n=1)^(\infty)u_n$ converges if and only if any of its remainders $r_n=\sum\limits_(k=n+1)^(\infty)u_k converges $. It follows from this that discarding or adding a finite number of terms to a certain series does not change the convergence of the series.
Let us be given a certain alternating series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$, and let the first condition of the Leibniz test be satisfied for this series, i.e. $\lim_(n\to(\infty))u_n=0$. However, the second condition, i.e. $u_n≥u_(n+1)$, is executed starting from a certain number $n_0\in(N)$. If $n_0=1$, then we get the usual formulation of the second condition of Leibniz’s criterion, therefore the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$ will converge. If $n_0>1$, then we split the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$ into two parts. In the first part, we select all those elements whose numbers are less than $n_0$:
$$ \sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n=\sum\limits_(n=1)^(n_0-1)(-1)^(n +1)u_n+\sum\limits_(n=n_0)^(\infty)(-1)^(n+1)u_n $$
For the series $\sum\limits_(n=n_0)^(\infty)(-1)^(n+1)u_n$ both conditions of the Leibniz test are satisfied, therefore the series $\sum\limits_(n=n_0)^(\ infty)(-1)^(n+1)u_n$ converges. Since the remainder converges, the original series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$ will also converge.
Thus, it does not matter at all whether the second condition of the Leibniz test is satisfied, starting from the first, or from the thousandth element - the series will still converge.
I note that Leibniz's test is sufficient, but not a necessary condition convergence of alternating series. In other words, the fulfillment of the conditions of the Leibniz criterion guarantees the convergence of the series, but failure to satisfy these conditions does not guarantee either convergence or divergence. Of course, failure to fulfill the first condition, i.e. case $\lim_(n\to(\infty))u_n\neq(0)$, means the divergence of the series $\sum\limits_(n=n_0)^(\infty)(-1)^(n+1)u_n $, however, failure to satisfy the second condition can occur for both convergent and divergent series.
Since alternating series of signs are often found in standard standard calculations, I have compiled a scheme by which a standard alternating series of signs can be examined for convergence.
Of course, you can directly apply the Leibniz test, bypassing the check for the convergence of a series of modules. However, for standard educational examples, checking a series of modules is necessary, since most authors of standard calculations require not only to find out whether the series converges or not, but to determine the nature of convergence (conditional or absolute). Let's move on to examples.
Example No. 1
Examine the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)\frac(4n-1)(n^2+3n)$ for convergence.
First, let's find out whether this series is really alternating. Since $n≥1$, then $4n-1≥3>0$ and $n^2+3n≥4>0$, i.e. for all $n\in(N)$ we have $\frac(4n-1)(n^2+3n)>0$. Thus, the given series has the form $\sum\limits_(n=1)^(\infty)(-1)^(n+1)u_n$, where $u_n=\frac(4n-1)(n^2 +3n)>0$, i.e. The series under consideration is alternating.
Usually such a check is done orally, but it is highly undesirable to skip it: errors in standard calculations are not uncommon. It often happens that the signs of the members of a given series begin to alternate not from the first member of the series. In this case, you can discard the “interfering” terms of the series and examine the convergence of the remainder (see the note at the beginning of this page).
So, we are given a sign-alternating series. We will follow the above. To begin with, let’s create a series of modules of members of this series:
$$ \sum\limits_(n=1)^(\infty)\left|(-1)^(n+1)\frac(4n-1)(n^2+3n)\right| =\sum\limits_(n=1)^(\infty)\frac(4n-1)(n^2+3n) $$
Let's check whether the compiled series of modules converges. Let's apply the comparison criterion. Since for all $n\in(N)$ we have $4n-1=3n+n-1≥3n$ and $n^2+3n≤n^2+3n^2=4n^2$, then:
$$ \frac(4n-1)(n^2+3n)≥ \frac(3n)(4n^2)=\frac(3)(4)\cdot\frac(1)(n) $$
The harmonic series $\sum\limits_(n=1)^(\infty)\frac(1)(n)$ diverges, so the series $\sum\limits_(n=1)^(\infty)\left will also diverge (\frac(3)(4)\cdot\frac(1)(n)\right)$. Therefore, according to the comparison criterion, the series $\sum\limits_(n=1)^(\infty)\frac(4n-1)(n^2+3n)$ diverges. Let us denote $u_n=\frac(4n-1)(n^2+3n)$ and check whether the conditions of the Leibniz test are satisfied for the original alternating series. Let's find $\lim_(n\to(\infty))u_n$:
$$ \lim_(n\to(\infty))u_n =\lim_(n\to(\infty))\frac(4n-1)(n^2+3n) =\lim_(n\to(\infty ))\frac(\frac(4)(n)-\frac(1)(n^2))(1+\frac(3)(n)) =0. $$
The first condition of Leibniz's test is satisfied. Now we need to find out whether the inequality $u_n≥u_(n+1)$ holds. A considerable number of authors prefer to write down the first few terms of the series and then conclude that the inequality $u_n≥u_(n+1)$ is satisfied.
In other words, this “proof” for this series would look like this: $\frac(2)(3)≤\frac(5)(8)≤\frac(8)(15)≤\ldots$. After comparing the first few terms, the conclusion is drawn: for the remaining terms the inequality will remain, each subsequent one will be no more than the previous one. I don’t know where this “method of proof” came from, but it is wrong. For example, for the sequence $v_n=\frac(10^n)(n$ получим такие первые члены: $v_1=10$, $v_2=50$, $v_3=\frac{500}{3}$, $v_4=\frac{1250}{3}$. Как видите, они возрастают, т.е., если ограничиться сравнением нескольких первых членов, то можно сделать вывод, что $v_{n+1}>v_n$ для всех $n\in{N}$. Однако такой вывод будет категорически неверным, так как начиная с $n=10$ элементы последовательности будут убывать.!}
How to prove the inequality $u_n≥u_(n+1)$? In general, there are several ways to do this. The simplest one in our case is to consider the difference $u_n-u_(n+1)$ and find out its sign. In the next example, we will consider a different method: by proving the decrease of the corresponding function.
$$ u_n-u_(n+1) =\frac(4n-1)(n^2+3n)-\frac(4(n+1)-1)((n+1)^2+3(n +1)) =\frac(4n-1)(n^2+3n)-\frac(4n+3)(n^2+5n+4)=\\ =\frac((4n-1)\cdot \left(n^2+5n+4\right)-\left(n^2+3n\right)\cdot(4n+3))(\left(n^2+3n\right)\cdot\left( n^2+5n+4\right)) =\frac(4n^2+2n-4)(\left(n^2+3n\right)\cdot\left(n^2+5n+4\right) ). $$
Since $n≥1$, then $4n^2-4≥0$, whence we have $4n^2+2n-4>0$, i.e. $u_n-u_(n+1)>0$, $u_n>u_(n+1)$. It happens, of course, that the inequality $u_n≥u_(n+1)$ is not satisfied from the first term of the series, but this is unimportant (see at the beginning of the page).
Thus, both conditions of the Leibniz criterion are satisfied. Since in this case the series $\sum\limits_(n=1)^(\infty)\left|(-1)^(n+1)\frac(4n-1)(n^2+3n)\right| $ diverges, then the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)\frac(4n-1)(n^2+3n)$ converges conditionally.
Answer: the series converges conditionally.
Example No. 2
Examine the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)\frac(5n-4)(\sqrt(2n^3-1))$ for convergence.
First, consider the expression $\frac(5n-4)(\sqrt(2n^3-1))$. It's worth doing a little checking to see if the condition is correct. The fact is that very often in the conditions of standard standard calculations one can encounter errors when the radical expression is negative, or a zero appears in the denominator for some values of $n$.
In order to avoid such troubles, let’s do a simple preliminary study. Since for $n≥1$ we have $2n^3≥2$, then $2n^3-1≥1$, i.e. the expression under the root cannot be negative or equal to zero. Therefore, the condition is quite correct. The expression $\frac(5n-4)(\sqrt(2n^3-1))$ is defined for all $n≥1$.
Let me add that for $n≥1$ the inequality $\frac(5n-4)(\sqrt(2n^3-1))>0$ is true, i.e. We are given a sign-alternating series. We will explore it according to the above. To begin with, let’s create a series of modules of members of this series:
$$ \sum\limits_(n=1)^(\infty)\left|(-1)^(n+1)\frac(5n-4)(\sqrt(2n^3-1))\right| =\sum\limits_(n=1)^(\infty)\frac(5n-4)(\sqrt(2n^3-1)) $$
Let's check whether a series composed of moduli of members of a given series converges. Let's apply the comparison criterion. In solving the previous example, we used the first comparison criterion. Here, purely for variety, we apply the second sign of comparison (the sign of comparison in the limiting form). Let us compare the series $\sum\limits_(n=1)^(\infty)\frac(5n-4)(\sqrt(2n^3-1))$ with the divergent series $\sum\limits_(n=1)^ (\infty)\frac(1)(\sqrt(n))$:
$$ \lim_(n\to\infty)\frac(\frac(5n-4)(\sqrt(2n^3-1)))(\frac(1)(\sqrt(n))) =\lim_ (n\to\infty)\frac(5n\sqrt(n)-4\sqrt(n))(\sqrt(2n^3-1)) =\lim_(n\to\infty)\frac(\frac (5n\sqrt(n))(n\sqrt(n))-\frac(4\sqrt(n))(n\sqrt(n)))(\sqrt(\frac(2n^3-1)( n^3))) \lim_(n\to\infty)\frac(5-\frac(4)(n))(\sqrt(2-\frac(1)(n^3))) =\frac (5)(\sqrt(2)). $$
Since $\frac(5)(\sqrt(2))\neq(0)$ and $\frac(5)(\sqrt(2))\neq\infty$, then simultaneously with the series $\sum\limits_ (n=1)^(\infty)\frac(1)(\sqrt(n))$ will diverge and the series $\sum\limits_(n=1)^(\infty)\frac(5n-4)( \sqrt(2n^3-1))$.
So, the given alternating series does not have absolute convergence. Let us denote $u_n=\frac(5n-4)(\sqrt(2n^3-1))$ and check whether the conditions of the Leibniz test are satisfied. Let's find $\lim_(n\to(\infty))u_n$:
$$ \lim_(n\to(\infty))u_n =\lim_(n\to(\infty))\frac(5n-4)(\sqrt(2n^3-1)) =\lim_(n\ to(\infty))\frac(\frac(5n)(n^(\frac(3)(2)))-\frac(4)(n^(\frac(3)(2))))( \sqrt(\frac(2n^3-1)(n^3))) =\lim_(n\to(\infty))\frac(\frac(5)(\sqrt(n))-\frac( 4)(n^(\frac(3)(2))))(\sqrt(2-\frac(1)(n^3))) =0. $$
The first condition of Leibniz's test is satisfied. Now we need to find out whether the inequality $u_n≥u_(n+1)$ holds. In the previous example, we looked at one of the ways to prove this inequality: by finding out the sign of the difference $u_n-u_(n+1)$. This time let's use a different method: instead of $u_n=\frac(5n-4)(\sqrt(2n^3-1))$, consider the function $y(x)=\frac(5x-4)(\sqrt( 2x^3-1))$ provided $x≥1$. I note that the behavior of this function under the condition $x<1$ нам совершенно безразлично.
Our goal is to prove that the function $y(x)$ is nonincreasing (or decreasing). If we prove that the function $y(x)$ is non-increasing, then for all values $x_2>x_1$ we will have $y(x_1)≥y(x_2)$. Assuming $x_1=n$ and $x_2=n+1$, we obtain that the inequality $n+1>n$ implies the truth of the inequality $y(n)≥y(n+1)$. Since $y(n)=u_n$, then the inequality $y(n)≥y(n+1)$ is the same as $u_(n)≥u_(n+1)$.
If we show that $y(x)$ is a decreasing function, then the inequality $n+1>n$ will lead to the truth of the inequality $y(n)>y(n+1)$, i.e. $u_(n)>u_(n+1)$.
Let's find the derivative $y"(x)$ and find out its sign for the corresponding values of $x$.
$$ y"(x)=\frac((5x-4)"\cdot\sqrt(2x^3-1)-(5x-4)\cdot\left(\sqrt(2x^3-1)\right )")(\left(\sqrt(2x^3-1)\right)^2) =\frac(5\cdot\sqrt(2x^3-1)-(5x-4)\cdot\frac(1 )(2\sqrt(2x^3-1))\cdot(6x^2))(2x^3-1)=\\ =\frac(5\cdot\left(2x^3-1\right)- (5x-4)\cdot(3x^2))(\left(2x^3-1\right)^(\frac(3)(2))) =\frac(-5x^3+12x^2- 5)(\left(2x^3-1\right)^(\frac(3)(2))) $$
I think it is obvious that for sufficiently large positive values of $x≥1$ the polynomial in the denominator will be less than zero, i.e. $-5x^3+12x^2-5<0$. Эту "очевидность" несложно обосновать формально - если вспомнить курс алгебры. Дело в том, что согласно лемме о модуле старшего члена, при достаточно больших значениях $|x|$ знак многочлена совпадает с знаком его старшего члена. Адаптируясь к нашей задаче получаем, что существует такое число $c≥1$, то для всех $x≥c$ будет верным неравенство $-5x^3+12x^2-5<0$. В принципе, существования такого числа $c$ уже вполне достаточно для дальнейшего решения задачи.
However, let's approach the issue less formally. In order not to involve unnecessary lemmas from algebra, we will simply roughly estimate the value of the expression $-5x^3+12x^2-5$. Let's take into account $-5x^3+12x^2-5=x^2(-5x+12)-5$. For $x≥3$ we have $-5x+12<0$, посему $x^2(-5x+12)-5<0$.
Thus, for $x≥3$ we have $y"(x)<0$, т.е. функция $y(x)$ убывает. А это, в свою очередь, означает, что при $n≥3$ верно неравенство $u_n>u_(n+1)$, i.e. the second condition of Leibniz's test is satisfied. Of course, we showed the fulfillment of the second condition not with $n=1$, but with $n=3$, but this is unimportant (see at the beginning of the page).
Thus, both conditions of the Leibniz criterion are satisfied. Since in this case the series $\sum\limits_(n=1)^(\infty)\left|(-1)^(n+1)\frac(5n-4)(\sqrt(2n^3-1) )\right|$ diverges, then the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)\frac(4n-1)(n^2+3n)$ converges conditionally.
Answer: the series converges conditionally.
Example No. 3
Examine the series $\sum\limits_(n=1)^(\infty)(-1)^(n+1)\frac(3n+4)(2^n)$ for convergence.
This example is not of great interest, so I will write it briefly. We are given an alternating series, which we will again explore using . Let's make a series of modules of the members of this series:
$$ \sum\limits_(n=1)^(\infty)\left|(-1)^(n+1)\frac(3n+4)(2^n)\right| =\sum\limits_(n=1)^(\infty)\frac(3n+4)(2^n) $$
Let's apply D'Alembert's sign. Denoting $u_n=\frac(3n+4)(2^n)$, we get $u_(n+1)=\frac(3n+7)(2^(n+1))$ .
$$ \lim_(n\to\infty)\frac(u_(n+1))(u_(n)) =\lim_(n\to\infty)\frac(\frac(3n+7)(2^ (n+1)))(\frac(3n+4)(2^n)) =\frac(1)(2)\lim_(n\to\infty)\frac(3n+7)(3n+4 ) =\frac(1)(2)\lim_(n\to\infty)\frac(3+\frac(7)(n))(3+\frac(4)(n)) =\frac(1 )(2)\cdot(1)=\frac(1)(2). $$
Since $\frac(1)(2)<1$, то согласно признаку Д"Аламбера ряд $\sum\limits_{n=1}^{\infty}\frac{3n+4}{2^n}$ сходится. Из сходимости ряда $\sum\limits_{n=1}^{\infty}\left|(-1)^{n+1}\frac{3n+4}{2^n}\right|$, что ряд $\sum\limits_{n=1}^{\infty}(-1)^{n+1}\frac{3n+4}{2^n}$ сходится, причём сходится абсолютно.
I note that to solve the given example we did not need the Leibniz test. That is why it is convenient to first check the convergence of a series of modules, and then, if necessary, investigate the convergence of the original alternating series.
Answer: the series converges absolutely.
Definition 1
The number series $\sum \limits _(n=1)^(\infty )u_(n) $, the terms of which have arbitrary signs (+), (?), is called an alternating series.
The alternating series discussed above are a special case of an alternating series; It is clear that not every alternating series is alternating. For example, the series $1-\frac(1)(2) -\frac(1)(3) +\frac(1)(4) +\frac(1)(5) -\frac(1)(6) - \frac(1)(7) +\ldots - $ alternating, but not an alternating series.
Note that in an alternating series there are infinitely many terms with both the sign (+) and the sign (-). If this is not true, for example, the series contains a finite number of negative terms, then they can be discarded and a series composed only of positive terms can be considered, and vice versa.
Definition 2
If the number series $\sum \limits _(n=1)^(\infty )u_(n) $ converges and its sum is equal to S, and the partial sum is equal to $S_n$ , then $r_(n) =S-S_( n) $ is called the remainder of the series, and $\mathop(\lim )\limits_(n\to \infty ) r_(n) =\mathop(\lim )\limits_(n\to \infty ) (S-S_(n ))=S-S=0$, i.e. the remainder of the convergent series tends to 0.
Definition 3
The series $\sum \limits _(n=1)^(\infty )u_(n) $ is called absolutely convergent if the series composed of the absolute values of its terms $\sum \limits _(n=1)^(\ infty )\left|u_(n) \right| $.
Definition 4
If the number series $\sum \limits _(n=1)^(\infty )u_(n) $ converges, and the series $\sum \limits _(n=1)^(\infty )\left|u_(n )\right| $, composed of the absolute values of its members, diverges, then the original series is called conditionally (non-absolutely) convergent.
Theorem 1 (a sufficient criterion for the convergence of alternating series)
An alternating series $\sum \limits _(n=1)^(\infty )u_(n) $ converges, and absolutely, if the series composed of the absolute values of its terms converges $\sum \limits _(n=1)^ (\infty )\left|u_(n) \right| $.
Comment
Theorem 1 provides only a sufficient condition for the convergence of alternating series. The converse theorem is not true, i.e. if the alternating series $\sum \limits _(n=1)^(\infty )u_(n) $ converges, then it is not necessary that the series composed of the modules $\sum \limits _(n=1)^( \infty )\left|u_(n) \right| $ (it can be either convergent or divergent). For example, the series $1-\frac(1)(2) +\frac(1)(3) -\frac(1)(4) +...=\sum \limits _(n=1)^(\infty )\frac((-1)^(n-1) )(n) $ converges according to Leibniz's criterion, and the series composed of the absolute values of its terms $\sum \limits _(n=1)^(\infty ) \, \frac(1)(n) $ (harmonic series) diverges.
Property 1
If the series $\sum \limits _(n=1)^(\infty )u_(n) $ is absolutely convergent, then it converges absolutely for any permutation of its terms, and the sum of the series does not depend on the order of the terms. If $S"$ is the sum of all its positive terms, and $S""$ is the sum of all absolute values of negative terms, then the sum of the series $\sum \limits _(n=1)^(\infty )u_(n) $ is equal to $S=S"-S""$.
Property 2
If the series $\sum \limits _(n=1)^(\infty )u_(n) $ is absolutely convergent and $C=(\rm const)$, then the series $\sum \limits _(n=1)^ (\infty )C\cdot u_(n) $ is also absolutely convergent.
Property 3
If the series $\sum \limits _(n=1)^(\infty )u_(n) $ and $\sum \limits _(n=1)^(\infty )v_(n) $ are absolutely convergent, then the series $\sum \limits _(n=1)^(\infty )(u_(n) \pm v_(n)) $ are also absolutely convergent.
Property 4 (Riemann's theorem)
If the series is conditionally convergent, then no matter what number A we take, we can rearrange the terms of this series so that its sum turns out to be exactly equal to A; Moreover, it is possible to rearrange the terms of a conditionally convergent series so that after this it diverges.
Example 1
Examine the series for conditional and absolute convergence
\[\sum \limits _(n=1)^(\infty )\frac((-1)^(n) \cdot 9^(n) )(n .\] !}
Solution. This series is alternating, the general term of which will be denoted by: $\frac((-1)^(n) \cdot 9^(n) )(n =u_{n} $. Составим ряд из абсолютных величин $\sum \limits _{n=1}^{\infty }\left|u_{n} \right| =\sum \limits _{n=1}^{\infty }\frac{9^{n} }{n!} $ и применим к нему признак Даламбера. Составим предел $\mathop{\lim }\limits_{n\to \infty } \frac{a_{n+1} }{a_{n} } $, где $a_{n} =\frac{9^{n} }{n!} $, $a_{n+1} =\frac{9^{n+1} }{(n+1)!} $. Проведя преобразования, получаем $\mathop{\lim }\limits_{n\to \infty } \frac{a_{n+1} }{a_{n} } =\mathop{\lim }\limits_{n\to \infty } \frac{9^{n+1} \cdot n!}{(n+1)!\cdot 9^{n} } =\mathop{\lim }\limits_{n\to \infty } \frac{9^{n} \cdot 9\cdot n!}{n!\cdot (n+1)\cdot 9^{n} } =\mathop{\lim }\limits_{n\to \infty } \frac{9}{n+1} =0$. Таким образом, ряд $\sum \limits _{n=1}^{\infty }\left|u_{n} \right| =\sum \limits _{n=1}^{\infty }\frac{9^{n} }{n!} $ сходится, а значит, исходный знакопеременный ряд сходится абсолютно.Ответ: ряд $\sum \limits _{n=1}^{\infty }\frac{(-1)^{n} \cdot 9^{n} }{n!} $ абсолютно сходится.!}
Example 2
Examine the series $\sum \limits _(n=1)^(\infty )\frac((-1)^(n) \cdot \sqrt(n) )(n+1) $ for absolute and conditional convergence.
- Let us examine the series for absolute convergence. Let us denote $\frac((-1)^(n) \cdot \sqrt(n) )(n+1) =u_(n) $ and compose a series of absolute values $a_(n) =\left|u_(n ) \right|=\frac(\sqrt(n) )(n+1) $. We get the series $\sum \limits _(n=1)^(\infty )\left|u_(n) \right| =\sum \limits _(n=1)^(\infty )\, \frac(\sqrt(n) )(n+1) $ with positive terms, to which we apply the limit test for comparing series. For comparison with the series $\sum \limits _(n=1)^(\infty )a_(n) =\sum \limits _(n=1)^(\infty )\, \frac(\sqrt(n) )(n+1) $ consider a series that has the form $\sum \limits _(n=1)^(\infty )\, b_(n) =\sum \limits _(n=1)^(\infty )\, \frac(1)(\sqrt(n) ) \, $. This series is a Dirichlet series with exponent $p=\frac(1)(2)
- Next, we examine the original series $\sum \limits _(n=1)^(\infty )\frac((-1)^(n) \cdot \sqrt(n) )(n+1) $ for conditional convergence. To do this, we check the fulfillment of the conditions of the Leibniz test. Condition 1): $u_(n) =(-1)^(n) \cdot a_(n) $, where $a_(n) =\frac(\sqrt(n) )(n+1) >0$ , i.e. this series is alternating. To check condition 2) about the monotonic decrease of the terms of the series, we use the following method. Consider the auxiliary function $f(x)=\frac(\sqrt(x) )(x+1) $ defined at $x\in . The expression in square brackets is positive and S2n>0, so S2n<u 1 for anyone n. Thus, the sequence of partial sums S2n increases and is limited, therefore there is a finite S2n=S. At the same time 0<S≤u 1.
Let us now consider the partial sum of an odd number of terms of the series S 2 n +1=S2n+u 2 n +1. Let us pass in the last equality to the limit at n→∞: S 2 n +1 = S 2 n + u 2 n +1 =S+ 0=S. Thus, the partial sums of both even and odd numbers of terms of the series have the same limit S, That's why S n=S, that is, this series converges. The theorem has been proven.
Example.
Examine the series for convergence
Let's apply Leibniz's test.
u n= >u n +1=
Both conditions of Leibniz's criterion are satisfied, therefore, the series converges.
Notes.
1. Leibniz’s theorem is valid even if the condition u n >u n + 1 is executed starting from some number N.
2. Condition u n >u n +1 is not necessary. The series may converge if it does not hold. For example, a series
converges as the difference of two convergent series although the condition u n >u n +1 is not executed.Definition 8. If an alternating series converges, but a series composed of the absolute values of the terms of this series diverges, then the alternating series is said to converge conditionally.
Definition 9. If both the alternating series itself and the series composed of the absolute values of its terms converge, then the alternating series is said to converge absolutely.
Example.
Establish the nature of convergence of the series
It is obvious that this series converges according to Leibniz's criterion. Indeed: and u n=
A series composed of absolute values of the terms of a given series is a divergent harmonic series. Therefore, this series converges conditionally.