Monotonic sequences

Decide whether the following sequences are monotonic and, if so, determine whether they are increasing, decreasing, non-decreasing or non-increasing.

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.