Alg9.3.4

From Exampleproblems

Find the sum of the infinite series $\frac{1}{1\cdot3}+\frac{1}{2}[\frac{1}{3\cdot5}]+\frac{1}{3}[\frac{1}{5\cdot7}].........\,$

nth term in the given series is $T_n=\frac{1}{n}[\frac{1}{(2n-1)(2n+1)}]=\frac{1}{(2n-1)(n)(2n+1)}=\frac{1}{2n-1}-\frac{1}{n}+\frac{1}{2n+1}\,$

Giving the values of 1,2,3.....upto n $T_1=\frac{1}{1}-\frac{1}{1}+\frac{1}{3}\,$

$T_2=\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\,$

$T_3=\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\,$

and so on... Adding all the above series $[\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+......]-[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....]+[\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......]-..............\,$

The above simplification results in $[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......]-1\,$

This results in $\mathrm{Log}_e(2)-1\,$

$\mathrm{Log}_e(2)-\mathrm{Log}_e(e)\,$

The solution is $\mathrm{Log}_e(\frac{2}{e})\,$