11The immediate jerk is to add 1 and subtract 1 from the numerators - from which a telescopic summation follows....
for x>1
evaluate
\frac{x}{x+1}+\frac{x^{2}}{(x+1)(x^{2}+1)}+\frac{x^{4}}{(x+1)(x^{2}+1)(x^{4}+1)}...........\infty
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4 Answers
11
11Basically what happens is this:-
\frac{x}{x+1}=1-\frac{1}{x+1}
\frac{x^2}{(x+1)(x^2+1)}=\frac{1}{x+1}-\frac{1}{(x+1)(x^2+1)}
So for n terms we have the sum as 1-\frac{1}{(x+1)(x^2+1)...(x^{2n-2}+1)}
As n goes to infinity, the left term vanishes, and we have the ans as 1.