Collection of Maths Problems

Monotonic sequences

Task number: 2862

• Variant 1

  $\left\{ 2n+(-1)^n \right\}_{n=1}^\infty$

• Resolution

  Let $a_n= 2n+(-1)^n$,

  For even $n=2k$ we have $a_{2k}=4k+1=2(2k+1)-1=a_{2k+1}$.

  For odd $n=2k+1$ we have $a_{2k+1}=4k+1<4k+5=a_{2k+2}$.

• Variant 2

  $\displaystyle \left\{ \frac1{1+n^2} \right\}_{n=1}^\infty$

• Resolution

  Let $a_n=\frac1{1+n^2}$.

  Then $\frac{a_n}{a_{n+1}}=\frac{1+(n+1)^2}{1+n^2}=1+\frac{2n+1}{1+n^2}>1$, so $a_n>a_{n+1}$.

• Result

  The sequence is decreasing.

• Variant 3

  $\displaystyle \left\{ \frac{n+1}{n+2} \right\}_{n=1}^\infty$

• Resolution

  Let $a_n=\frac{n+1}{n+2}$.

  Then $\frac{a_n}{a_{n-1}}=\frac{(n+1)^2}{n^2+2n}=1+\frac1{n^2+2n}>1$, so $a_n>a_{n-1}$.

• Result

  The sequence is increasing.

• Variant 4

  $\displaystyle \left\{ \frac{n+1}{\sqrt{n^2+2n-2}} \right\}_{n=1}^\infty$

• Resolution

  Let $a_n=\frac{n+1}{\sqrt{n^2+2n-2}}=\sqrt{1+\frac{3}{(n+1)^2-3}}$.

  Then for $n\ge 2$ we have $a_{n-1}^2-a_n^2=\frac{3}{n^2-3}-\frac{3}{(n+1)^2-3}=\frac{3(2n+1)}{(n^2-3)((n+1)^2-3)}>0$.

  From this we see that $a_{n-1}^2>a_n^2$ and by taking the square root we get $a_{n-1}>a_n$.

• Result

  The sequence is decreasing.