Введение в математический анализ в вопросах и задачах. Анчиков А.М - 9 стр.

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k + 1 x > −1.
∀n ∈ N x > −1.
S
n
= −1 + 2 − 3 + 4 − 5 + ... + (−1)
n
n. (6)
S
1
, S
2
, ...S
6
: S
1
= −1, S
2
=
−1 + 2 = 1, S
3
= S
2
− 3 = −2, S
4
= S
3
+ 4 = 2, S
5
= S
4
− 5 =
−3, S
6
= S − 5 + 6 = 3.
1 =
·
1 + 1
2
¸
=
·
2 + 1
2
¸
, 2 =
·
3 + 1
2
¸
=
·
4 + 1
2
¸
,
3 =
·
5 + 1
2
¸
=
·
6 + 1
2
¸
.
[a] a
S
n
= (−1)
n
·
n + 1
2
¸
. (7)
1, 2, ... 6
∀k > 6
S
k
= (−1)
k
"
k + 1
2
#
. (8)
S
k+1
= S
k
+ (−1)
k+1
(k + 1) = (−1)
k
"
k + 1
2
#
+ (−1)
k
(k + 1) =
= (−1)
k+1
Ã
k + 1 −
"
k + 1
2
#!
.
∀n ∈ N, [
n
2
] + [
n+1
2
] = n.
Ýòèì äîêàçàíî, ÷òî (5) ñïðàâåäëèâî äëÿ íàòóðàëüíîãî ÷èñëà
k + 1 è x > −1. Òåì ñàìûì äîêàçàíî, ÷òî (5) ñïðàâåäëèâî ïðè
∀n ∈ N è x > −1.
   Ïðèìåð 3. Íàéòè ñóììó
             Sn = −1 + 2 − 3 + 4 − 5 + ... + (−1)n n.                          (6)

   Ðåøåíèå. Ðàññìîòðèì S1 , S2 , ...S6 :    S1 = −1, S2 =
−1 + 2 = 1, S3 = S2 − 3 = −2, S4 = S3 + 4 = 2, S5 = S4 − 5 =
−3, S6 = S − 5 + 6 = 3. Ñ äðóãîé ñòîðîíû:
             ·      ¸     ·          ¸            ·           ¸        ·   ¸
              1+1     2+1            3+1         4+1
        1=         =         , 2=              =     ,
               2         2              2         2
                       ·     ¸    ·        ¸
                         5+1        6+1
                   3=          =             .
                           2          2
Çäåñü ïîä [a] ïîíèìàåòñÿ öåëàÿ ÷àñòü ÷èñëà a . Îòñþäà èìååì
ãèïîòåçó:                       ·        ¸
                                  n+1
                    Sn = (−1)n             .             (7)
                                    2
Äëÿ íàòóðàëüíûõ çíà÷åíèé 1, 2, ... 6 ñîîòíîøåíèå (7)ñïðàâåä-
ëèâî.
   Ïðåäïîëîæèì, ÷òî ∀k > 6 ñîîòíîøåíèå (7)ñïðàâåäëèâî:
                                             "            #
                                         k       k+1
                         Sk = (−1)                   .                         (8)
                                                  2
Äàëåå
                                                     "        #
                   k+1                           k       k+1
 Sk+1 = Sk + (−1)        (k + 1) = (−1)                      + (−1)k (k + 1) =
                                                          2
                                 Ã                   "        #!
                           k+1        k+1
                 = (−1)          k+1−                              .
                                       2
Çàìåòèì, ÷òî äëÿ ∀n ∈ N, [ n2 ] + [ n+1
                                     2
                                        ] = n. Èñïîëüçóÿ ïðåäû-

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