Question

Доказать методом математической индукции.

Answer 1

1) Пусть n=2
$1+ \frac{1}{ \sqrt{2} } \ \textgreater \ \sqrt{2} \\ \\ \sqrt{2}* (1+ \frac{1}{ \sqrt{2} } ) \ \textgreater \ \sqrt{2} *\sqrt{2} \\ \\ \sqrt{2} +1\ \textgreater \ 2 \\ \\ \sqrt{2} \ \textgreater \ 1$
верно
2)Пусть верно при n=k
$1+ \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{3} } +...+ \frac{1}{ \sqrt{k} } \ \textgreater \ \sqrt{k}$
3)докажем, что верно при n=k+1$1+ \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{3} } +...+ \frac{1}{ \sqrt{k} } + \frac{1}{ \sqrt{k+1} } \ \textgreater \ \sqrt{k}+ \frac{1}{ \sqrt{k+1} }$
$\frac{1}{ \sqrt{k+1} } -$положительное число
$\sqrt{k} + \frac{1}{ \sqrt{k+1} } \ \textgreater \ \sqrt{k+1} \\ \\ \sqrt{k+1}*( \sqrt{k} + \frac{1}{ \sqrt{k+1} } )\ \textgreater \ \sqrt{k+1} * \sqrt{k+1} \\ \\ \sqrt{k(k+1)} +1\ \textgreater \ k+1 \\ \\ \sqrt{k^2+k} \ \textgreater \ \sqrt{k^2} ;k \geq 2$
верно
⇒
$1+ \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{3} } +...+ \frac{1}{ \sqrt{k} } + \frac{1}{ \sqrt{k+1} } \ \textgreater \ \sqrt{k+1} }$
ч.т.д.