![\[
\left| {2 + i} \right| = \sqrt 5 > 2 \Rightarrow n(\frac{{\left| {2 + i} \right|}}{2})^n \to \infty \]](https://dxdy-04.korotkov.co.uk/f/3/5/1/3514e30bee79677c807d311f18232e5f82.png "\[
\left| {2 + i} \right| = \sqrt 5 > 2 \Rightarrow n(\frac{{\left| {2 + i} \right|}}{2})^n \to \infty \]")
Поэтому не выполняется необходимый признак сходимости ряда, и ряд не может сходиться, ни абсолютно, ни условно - никак не может сходиться.
emilj писал(а):
P.s.: Я знаю, что я уже надоел.
Надоели не Вы конкретно, а надоели вот такие диалоги: сначала я пишу человеку
Brukvalub писал(а):
Проверьте необходимое условие сходимости.
Brukvalub писал(а):
Нужно доказать, что модуль общего члена ряда не стремится к нулю, при этом член нужно брать весь, ничего не отбрасывая.
, а в ответ читаю:
emilj писал(а):
А если докажем, то что ээто значит? Как определить он условно или абсолютно сходится?
Согласитесь, создаётся впечатление, что человек
ничего не учил, да и подумать не хочет.
27.11.2007, 17:52 
![\[
\sum\limits_{n = 1}^\infty {\frac{{(2 + i)^n n }}
{2^n}}
\]](https://dxdy-01.korotkov.co.uk/f/0/1/e/01e1a5e813f8527400be836e728fe45882.png "\[
\sum\limits_{n = 1}^\infty {\frac{{(2 + i)^n n }}
{2^n}}
\]")
, как в знакочередующихся рядах? Если без, то выполняется условие! Покажите, пожалуйста, на подобном примере,если на этом нельзя, сам принцип исследования.
исследовать легко. Я просто хочу понять, ряд вида ![\[
\sum\limits_{n = 1}^\infty {|{\frac{{sin(n) }}
{n}} |}
\]](https://dxdy-02.korotkov.co.uk/f/5/d/8/5d827637676ffbb6eb9cc9aeea1f710582.png "\[
\sum\limits_{n = 1}^\infty {|{\frac{{sin(n) }}
{n}} |}
\]")
![\[
\sum\limits_{n = 1}^\infty {\frac{{1}} {{(n+i)^n} {n^(1/2)}}
\]](https://dxdy-01.korotkov.co.uk/f/0/0/2/0021d4ab8b187707873dffe5ab8ae8f782.png "\[
\sum\limits_{n = 1}^\infty {\frac{{1}} {{(n+i)^n} {n^(1/2)}}
\]")
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\[
\begin{gathered}
\sum\limits_{n = 1}^\infty {\frac{1}
{{(n + i)^n \sqrt n }}} \hfill \\
1)\mathop {\lim }\limits_{n \to \infty } |\frac{1}
{{(n + i)^n \sqrt n }}| = 0 \hfill \\
2) $Исследуем ряд на сходимость (т.к. исследуем на абсллютную сх-ть, то берем ряд по модулю)$\\ \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{1}
{{(n + i)^n \sqrt n }}} | \hfill \\
K = \mathop {\lim }\limits_{n \to \infty } \sqrt {|\frac{1}
{{(n + i)^n \sqrt n }}|} = \mathop {\lim }\limits_{n \to \infty } \frac{1}
{{|(n + i)n^{\frac{1}
{{2n}}} |}} = ... = \frac{1}
{\infty } = 0 \hfill \\
$Следовательно ряд сходится условно?$ \\
\end{gathered}
\]](https://dxdy-02.korotkov.co.uk/f/5/0/3/503837f6a3986dba8f9c9c3f94351f6782.png "% MathType!MTEF!2!1!+-
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\[
\begin{gathered}
\sum\limits_{n = 1}^\infty {\frac{1}
{{(n + i)^n \sqrt n }}} \hfill \\
1)\mathop {\lim }\limits_{n \to \infty } |\frac{1}
{{(n + i)^n \sqrt n }}| = 0 \hfill \\
2) $Исследуем ряд на сходимость (т.к. исследуем на абсллютную сх-ть, то берем ряд по модулю)$\\ \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{1}
{{(n + i)^n \sqrt n }}} | \hfill \\
K = \mathop {\lim }\limits_{n \to \infty } \sqrt {|\frac{1}
{{(n + i)^n \sqrt n }}|} = \mathop {\lim }\limits_{n \to \infty } \frac{1}
{{|(n + i)n^{\frac{1}
{{2n}}} |}} = ... = \frac{1}
{\infty } = 0 \hfill \\
$Следовательно ряд сходится условно?$ \\
\end{gathered}
\]")
![% MathType!MTEF!2!1!+-
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\[
\begin{gathered}
\sum\limits_{n = 1}^\infty {\frac{1}
{{(n + i)^n \sqrt n }} = } \left[ {|(n + i)|^{ - n} = \sqrt {n^2 + 1} e^{ - iarctg(\frac{1}
{n})} = \sqrt {n^2 + 1} (\cos (arctg(\frac{1}
{n})) - i\sin (arctg(\frac{1}
{n})))} \right] = \hfill \\
= \sum\limits_{n = 1}^\infty {\frac{n}
{{\sqrt n }} - i} \sum\limits_{n = 1}^\infty {\frac{1}
{{\sqrt n }}} \hfill \\
1)\mathop {\lim }\limits_{n \to \infty } \frac{1}
{{\sqrt n }} = 0 \hfill \\
2) \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{n}
{{\sqrt n }}|} = \sum\limits_{n = 1}^\infty {|\sqrt n |} \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{1}
{{\sqrt n }}} | \hfill \\
\hfill \\
\end{gathered}
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\[
\begin{gathered}
\sum\limits_{n = 1}^\infty {\frac{1}
{{(n + i)^n \sqrt n }} = } \left[ {|(n + i)|^{ - n} = \sqrt {n^2 + 1} e^{ - iarctg(\frac{1}
{n})} = \sqrt {n^2 + 1} (\cos (arctg(\frac{1}
{n})) - i\sin (arctg(\frac{1}
{n})))} \right] = \hfill \\
= \sum\limits_{n = 1}^\infty {\frac{n}
{{\sqrt n }} - i} \sum\limits_{n = 1}^\infty {\frac{1}
{{\sqrt n }}} \hfill \\
1)\mathop {\lim }\limits_{n \to \infty } \frac{1}
{{\sqrt n }} = 0 \hfill \\
2) \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{n}
{{\sqrt n }}|} = \sum\limits_{n = 1}^\infty {|\sqrt n |} \hfill \\
\sum\limits_{n = 1}^\infty {|\frac{1}
{{\sqrt n }}} | \hfill \\
\hfill \\
\end{gathered}
\]")
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\[
\sum\limits_{n = 1}^\infty {\frac{{\sin (\frac{1}
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{{\sqrt n }}}
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\[
\sum\limits_{n = 1}^\infty {\frac{{\sin (\frac{1}
{n}))}}
{{\sqrt n }}}
\]")
![\[ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{1}{{(n + i)^n \sqrt n }}} \\ 1)\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(n + i)^n \sqrt n }} = 0 \\ 2) $Исследуем ряд на сходимость (т.к. исследуем на абсллютную сх-ть, то берем ряд по модулю)$\\ \sum\limits_{n = 1}^\infty {\frac{1}{{(n + i)^n \sqrt n }}} \\ K = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{1}{{(n + i)^n \sqrt n }}} }} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(n + i)n^{\frac{1}{{2n}}} }}}} = ... = \frac{1}{0} = \infty \\ $Следовательно ряд сходится условно?$ \\ \end{array} \]](https://dxdy-01.korotkov.co.uk/f/c/1/e/c1e8c182844f5b45bd6bce0ffaabbc8782.png "\[ \begin{array}{l} \sum\limits_{n = 1}^\infty {\frac{1}{{(n + i)^n \sqrt n }}} \\ 1)\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(n + i)^n \sqrt n }} = 0 \\ 2) $Исследуем ряд на сходимость (т.к. исследуем на абсллютную сх-ть, то берем ряд по модулю)$\\ \sum\limits_{n = 1}^\infty {\frac{1}{{(n + i)^n \sqrt n }}} \\ K = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \sqrt {\frac{1}{{(n + i)^n \sqrt n }}} }} = \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \frac{1}{{(n + i)n^{\frac{1}{{2n}}} }}}} = ... = \frac{1}{0} = \infty \\ $Следовательно ряд сходится условно?$ \\ \end{array} \]")
![\[ \sum\limits_{n = 1}^\infty {\frac{{\sin (\frac{1} {n}))}} {{\sqrt n }}} \]](https://dxdy-01.korotkov.co.uk/f/8/6/3/8631720dbd7672a161c5669492e8b4a582.png "\[ \sum\limits_{n = 1}^\infty {\frac{{\sin (\frac{1} {n}))}} {{\sqrt n }}} \]")
Этот ряд нужно исследовать по признаку сравнения.