∑n=1∞1(x+n)(x+n+1)=∑n=1∞(1x+n−1x+n+1)=\sum^{\infty}_{n=1}\dfrac1{(x+n)(x+n+1)}=\sum^{\infty}_{n=1}(\dfrac1{x+n}-\dfrac1{x+n+1})=∑n=1∞(x+n)(x+n+1)1=∑n=1∞(x+n1−x+n+11)=
=limN→∞∑n=1N(1x+n−1x+n+1)=limN→∞(1x+1−1x+N+1)==\lim_{N\rightarrow \infty}\sum^{N}_{n=1}(\dfrac1{x+n}-\dfrac1{x+n+1})= \lim_{N\rightarrow \infty}(\dfrac1{x+1}-\dfrac1{x+N+1})==limN→∞∑n=1N(x+n1−x+n+11)=limN→∞(x+11−x+N+11)=
=1x+1−limN→∞1x+N+1=1x+1=\dfrac1{x+1}-\lim_{N\rightarrow \infty}\dfrac1{x+N+1}=\dfrac1{x+1}=x+11−limN→∞x+N+11=x+11
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