# if tn=n/(n+1)!  then Σ tn is equal to (where summation is from n=1 to n=20)

Dear student
$\sum _{\mathrm{n}=1}^{20}{\mathrm{t}}_{\mathrm{n}}\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{n}=1}^{20}\frac{\mathrm{n}}{\left(\mathrm{n}+1\right)!}\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{n}=1}^{20}\frac{\left(\mathrm{n}+1-1\right)}{\left(\mathrm{n}+1\right)!}\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{n}=1}^{20}\frac{\mathrm{n}+1}{\left(\mathrm{n}+1\right)!}-\frac{1}{\left(\mathrm{n}+1\right)!}\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{n}=1}^{20}\frac{\mathrm{n}+1}{\left(\mathrm{n}+1\right)\mathrm{n}!}-\frac{1}{\left(\mathrm{n}+1\right)!}\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{n}=1}^{20}\frac{1}{\mathrm{n}!}-\frac{1}{\left(\mathrm{n}+1\right)!}\phantom{\rule{0ex}{0ex}}=\frac{1}{1}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{19!}-\frac{1}{20!}+\frac{1}{20!}-\frac{1}{21!}\phantom{\rule{0ex}{0ex}}=1-\frac{1}{21!}\phantom{\rule{0ex}{0ex}}\approx 1$
Regards

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