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信息发布者：

$-\frac {a}{a+1}$

解：$(1)$∵$S_{1}=\frac 1a，$∴$S_{2}=-S_{1}-1=-\frac 1a-1=-\frac {a+1}a$

∴$S_{3}=\frac 1{S_{2}}=-\frac {a}{a+1}$

$(3)$∵$S_{1}=\frac 1{a}$

∴$S_{2}=-S_{1}-1=-\frac 1{a}-1=-\frac {a+1}{a}$

∴$S_{3}=\frac 1{S_{2}}=-\frac {a}{a+1}$

∴$S_{4}=-S_{3}-1=\frac {a}{a+1}-1=\frac {a-a-1}{a+1}=-\frac 1{a+1}$

∴$S_{5}=\frac 1{S_{4}}=-(a+1)$

∴$S_{6}=-S_{5}-1=a+1-1=a$

∴$S_{7}=\frac 1{S_{6}}=\frac 1{a}$

······

∴$S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6}$

$=\frac 1{a}+(-\frac {a+1}{a})+ (-\frac {a}{a+1})+(-\frac 1{a+1})+[-(a+1)]+a=-3$

∵$2022÷6=337$

∴$S_{1}+S_{2}+S_{3}+···+S_{2022}=(-3)×337=-1011$

$\frac {x+n+1}{x+n+2}-\frac {x+n}{x+n+1}=\frac 1{x+n+1}-\frac 1{x+n+2}$

证明：$(3)\frac {x+n+1}{x+n+2}-\frac {x+n}{x+n+1}$

$=\frac {(x+n+2)-1}{x+n+2}-\frac {(x+n+1)-1}{x+n+1}$

$=1-\frac 1{x+n+2}-(1-\frac 1{x+n+1})$

$=1-\frac 1{x+n+2}-1+\frac 1{x+n+1}$

$=\frac 1{x+n+1}-\frac 1{x+n+2}$

即$\frac {x+n+1}{x+n+2}-\frac {x+n}{x+n+1}=\frac 1{x+n+1}-\frac 1{x+n+2}$

$\frac {x+6}{x+7}-\frac {x+5}{x+6}=\frac 1{x+6}-\frac 1{x+7}$