[TEX]VT = \frac{{1!}}{m} + \frac{{2!}}{{m\left( {m + 1} \right)}} + \frac{{3!}}{{m\left( {m + 1} \right)\left( {m + 2} \right)}} + ... + \frac{{\left( {n + 1} \right)!}}{{m\left( {m + 1} \right)..\left( {m + n} \right)}}[/TEX]
do đó
[TEX]\begin{array}{l}VP - VT = \left( {\frac{{1!}}{{m - 2}} - \frac{{1!}}{m}} \right) - \frac{{2!}}{{m\left( {m + 1} \right)}} - \frac{{3!}}{{m\left( {m + 1} \right)\left( {m + 2} \right)}} - ... - \frac{{\left( {n + 1} \right)!}}{{m\left( {m + 1} \right)..\left( {m + n} \right)}}\\ = \left( {\frac{{2!}}{{\left( {m - 2} \right)m}} - \frac{{2!}}{{m\left( {m + 1} \right)}}} \right) - \frac{{3!}}{{m\left( {m + 1} \right)\left( {m + 2} \right)}} - ... - \frac{{\left( {n + 1} \right)!}}{{m\left( {m + 1} \right)..\left( {m + n} \right)}}\\ = \left( {\frac{{3!}}{{\left( {m - 2} \right)m\left( {m + 1} \right)}} - \frac{{3!}}{{m\left( {m + 1} \right)\left( {m + 2} \right)}}} \right) - ... - \frac{{\left( {n + 1} \right)!}}{{m\left( {m + 1} \right)..\left( {m + n} \right)}}\\ = \frac{{\left( {n + 2} \right)!}}{{\left( {m - 2} \right)m\left( {m + 1} \right)...\left( {m + n} \right)}} > 0\end{array}[/TEX]
=>dpcm
từ đó suy ra
[TEX]{\lim }\limits_{n - > + \infty } VT = \frac{1}{{m - 2}}[/TEX]
suy ra nếu đánh giá VT<a<VP thì sẽ sai
Chứng minh rằng [TEX]\forall k:=\overline{0;n}\ \ ; \ \ n\in Z^+[/TEX] ta luôn có
[TEX]\frac{1}{C_{2012}^1}+\frac{1}{C_{2013}^2}+.......+\frac{1}{C_{2012+k}^{k+1}}......+\frac{1}{C_{2012+n}^{n+1}}< \frac{1}{2010}[/TEX]
[TEX]C_{2012+n}^{n+1}= \frac{(2012+n)!}{(n+1)!1011!}=\frac{(n+2)(n+3)..(n+2012)}{1011!}[/TEX]
[TEX]\righ VT:=1011!\( \frac{1}{2.3.4.5...2012}+ \frac{1}{3.4.5...2013}+\frac{1}{4.5...2014}+...+ \frac{1}{(n+2)(n+3)..(n+2012)}\) [/TEX]
[TEX]=\( \frac{1!}{2012}+ \frac{2!}{2012.2013}+\frac{3!}{2012.1013.2014}+...+ \frac{n!}{2012.2013.2013...(n+2012)}\) [/TEX]
vodichhocmai said:
Rồi rồi lên bài trên nerversaynever.
Anh vodichhocmai. Thank em . Anh thấy ngay chỗ đó rồi
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