
\magnification 1200

�ada m��e b�t omezena dv�ma integr�ly, jeden
$$\int _{n+1}^{2n} 1/x \, dx , $$ druh�
$$\int _{n}^{2n-1} 1/x  \, dx . $$
Integr�ly se rovnaj� $\log(2n/(n+1))$ a $\log((2n-1)/n)$ s limitou
$\log 2$ pro $n \to \infty$. Proto je sou�et roven $\log 2$.

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{\bf Dal�� �e�en� : }
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a) prvni odhad :-)
   $$\sum_{k = 1}^n {1 \over k + n} <
       \sum_{k = 1}^n {1 \over n} = 1$$
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b) vyjadrime radu pro $n = k$ a $n = k + 1$

   ba) $n = k:$
   $$A = {1 \over k + 1} + \ldots  + {1 \over 2k}$$

   bb) $n = k + 1:$
   $$B = {1 \over k + 2} + \ldots + {1 \over 2k}
          + {1 \over 2k + 1} + {1 \over 2k + 2}$$

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c) udelame jejich rozdil

   $$B - A = {1 \over k + 2} + {1 \over 2k + 1}
             - {1 \over k + 1}
                      = {4k + 3 \over (2k + 1)(2k + 2)}
- {1 \over k + 1} = $$ $$
      = {(4k + 3)(k + 1) - (2k + 1)(2k + 2) \over
(2k + 1)(2k + 2)(k + 1)}
$$

$$
      = {4k^2 + 7k + 3 - 4k^2 - 6k - 2 \over
(2k + 1)(2k + 2)(k + 1)}
      = {1 \over (2k + 1)(2k + 2)}$$

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d) limita pro $n \to \infty$ z rady, kterou se ma secist, je
souctem rady pro $n = 1$ a prislusnych rozdilu od $1$ do $\infty$
$$ S = {1 \over 2} + \sum_{k = 1}^\infty
       \left( {1 \over (2k + 1)(2k + 2)} \right)
     = \sum_{k = 0}^\infty
       \left( {1 \over (2k + 1)(2k + 2)} \right) = $$ $$
     = \sum_{k = 0}^\infty
       \left( {1 \over 2k + 1} - {1 \over 2k + 2} \right)
     = \sum_{i = 1}^\infty {(-1)^{n+1} \over n} = \ln 2$$


\bye
