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\begin{itemize}

  \item
  dla każdego n $\in N_+$ \[\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}\]
  \item KB: dla n=1\\~\\
  L: $\frac{1}{i(i+1)} = \frac{1}{1(1+1)} = \frac{1}{2}$\\
  P: $\frac{n}{n+1} = \frac{1}{1+1} = \frac{1}{2}$\\~\\
  L = P
  \item Ki: Jeśli
  $1+...+\frac{1}{i(i+1)} = \frac{n}{n+1}$ to $\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$\\
  1+...+$\frac{1}{n(n+1)}=\frac{n}{n+1}$\\~\\
  P: $\frac{n+1}{n+2}$\\
  L: $\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)^\cancel{2}}{(n+2)\cancel{(n+1)}} = \frac{(n+1)}{(n+2)}$
  \begin{flushright}
  L = P $\square$
  \end{flushright}
\end{itemize}

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