![]() The integral does not converge, and so the series does not converge either. #= 2(oo)^("1/2") - cancel(2(1+1)^("1/2"))^"small"# Here we consider the limit of the function f (x)1/x as x approaches 0, and as x approaches infinity. Basically, it's a half-open integral that extends its domain forever and so it has no finite area. So, essentially, we have to integrate this, which is indeed continuous, positive, and decreasing at #[1,oo)#:Īt this point we know that #sqrt(x+1)# is a constantly increasing function, so it has an "open" accumulation that can never stop without a well-defined right-end boundary. There exists another function #f(x)# that is continuous, positive, and decreasing such that the convergence or divergence of #int_k^(oo)f(x)dx# determines the convergence or divergence of #sum_(n=k)^(oo)a_n#, respectively. Let there be a function #f(n) = a_n# where #a_n# is a series lying within the domain #[k,oo)#. Is infinity divided by infinity 1 Is infinity divided by infinity equal to 1 No. The integral test basically works from the definition of the integral (quick version: the integral is the accumulated sum of infinitely thin differential intervals #dn# over a specified interval #a->b#).Ī paraphrased version of the integral test is as follows: What is 1 divided by infinity squared Infinity is an undetermined or undefined value, it can very big and dividing 1 by that number will result to the value close to 0. ![]()
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