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![f(x)=ln\left[1-\dfrac{2}{(x+1)(x+2)}\right]=ln\left[\dfrac{(x+1)(x+2)-2}{(x+1)(x+2)}\right]=\\\\ln\left[\dfrac{x^2+3x}{(x+1)(x+2)}\right]=ln\left[\dfrac{x(x+3)}{(x+1)(x+2)}\right]=ln[x(x+3)]-ln[(x+1)(x+2)=\\\\=lnx+ln(x+3)-ln(x+1)-ln(x+2);\\f(k)=lnk+ln(k+3)-ln(k+1)-ln(k+2)=\\=lnk-ln(k+1)+ln(k+3)-ln(k+2),\ deci:\\\\a_n=\sum\limits_{k=1}^nf(k)=\\\\=ln1-ln2+ln4-ln3+\\+ln2-ln3+ln5-ln4+\\+ln3-ln4+ln6-ln5+\\+ln4-ln5+ln7-ln6+\\+\ldots+\\+ln(n-1)-lnn+ln(n+2)-ln(n+1)+\\+lnn-ln(n+1)+ln(n+3)-ln(n+2)=ln1-ln(n+1)+ln(n+3)-ln3=\\\\=ln\left(\dfrac{n+3}{n+1}\right)-ln3\to\ ln1-ln3=-ln3.](https://tex.z-dn.net/?f=f%28x%29%3Dln%5Cleft%5B1-%5Cdfrac%7B2%7D%7B%28x%2B1%29%28x%2B2%29%7D%5Cright%5D%3Dln%5Cleft%5B%5Cdfrac%7B%28x%2B1%29%28x%2B2%29-2%7D%7B%28x%2B1%29%28x%2B2%29%7D%5Cright%5D%3D%5C%5C%5C%5Cln%5Cleft%5B%5Cdfrac%7Bx%5E2%2B3x%7D%7B%28x%2B1%29%28x%2B2%29%7D%5Cright%5D%3Dln%5Cleft%5B%5Cdfrac%7Bx%28x%2B3%29%7D%7B%28x%2B1%29%28x%2B2%29%7D%5Cright%5D%3Dln%5Bx%28x%2B3%29%5D-ln%5B%28x%2B1%29%28x%2B2%29%3D%5C%5C%5C%5C%3Dlnx%2Bln%28x%2B3%29-ln%28x%2B1%29-ln%28x%2B2%29%3B%5C%5Cf%28k%29%3Dlnk%2Bln%28k%2B3%29-ln%28k%2B1%29-ln%28k%2B2%29%3D%5C%5C%3Dlnk-ln%28k%2B1%29%2Bln%28k%2B3%29-ln%28k%2B2%29%2C%5C+deci%3A%5C%5C%5C%5Ca_n%3D%5Csum%5Climits_%7Bk%3D1%7D%5Enf%28k%29%3D%5C%5C%5C%5C%3Dln1-ln2%2Bln4-ln3%2B%5C%5C%2Bln2-ln3%2Bln5-ln4%2B%5C%5C%2Bln3-ln4%2Bln6-ln5%2B%5C%5C%2Bln4-ln5%2Bln7-ln6%2B%5C%5C%2B%5Cldots%2B%5C%5C%2Bln%28n-1%29-lnn%2Bln%28n%2B2%29-ln%28n%2B1%29%2B%5C%5C%2Blnn-ln%28n%2B1%29%2Bln%28n%2B3%29-ln%28n%2B2%29%3Dln1-ln%28n%2B1%29%2Bln%28n%2B3%29-ln3%3D%5C%5C%5C%5C%3Dln%5Cleft%28%5Cdfrac%7Bn%2B3%7D%7Bn%2B1%7D%5Cright%29-ln3%5Cto%5C+ln1-ln3%3D-ln3. "f(x)=ln\left[1-\dfrac{2}{(x+1)(x+2)}\right]=ln\left[\dfrac{(x+1)(x+2)-2}{(x+1)(x+2)}\right]=\\\\ln\left[\dfrac{x^2+3x}{(x+1)(x+2)}\right]=ln\left[\dfrac{x(x+3)}{(x+1)(x+2)}\right]=ln[x(x+3)]-ln[(x+1)(x+2)=\\\\=lnx+ln(x+3)-ln(x+1)-ln(x+2);\\f(k)=lnk+ln(k+3)-ln(k+1)-ln(k+2)=\\=lnk-ln(k+1)+ln(k+3)-ln(k+2),\ deci:\\\\a_n=\sum\limits_{k=1}^nf(k)=\\\\=ln1-ln2+ln4-ln3+\\+ln2-ln3+ln5-ln4+\\+ln3-ln4+ln6-ln5+\\+ln4-ln5+ln7-ln6+\\+\ldots+\\+ln(n-1)-lnn+ln(n+2)-ln(n+1)+\\+lnn-ln(n+1)+ln(n+3)-ln(n+2)=ln1-ln(n+1)+ln(n+3)-ln3=\\\\=ln\left(\dfrac{n+3}{n+1}\right)-ln3\to\ ln1-ln3=-ln3.")
Green eyes.
Green eyes.
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