How do you use the integral test to determine if #1/2+1/5+1/10+1/17+1/26+...# is convergent or divergent?

Question

Emily Ammerman · Accepted Answer

The series: #sum_(n=1)^oo 1/(n^2+1)# is convergent.

The series is: #sum_(n=1)^oo 1/(n^2+1)# so we can use for the integral test the function: #f(x) = 1/(1+x^2)# which is positive, decreasing and infinitesimal for #x->oo#, thus satisfying all the hypotheses of the integral test. Calculating the integral: #int_1^oo 1/(1+x^2)dx = [arctan x]_1^oo = pi/2-pi/4 =pi/4# So the series is convergent.

Emma Aldridge · Answer

undefined To use the integral test to determine if the series 1/2 + 1/5 + 1/10 + 1/17 + 1/26 + ... is convergent or divergent, we first need to express the series as a function of x in terms of an integral. Notice that the terms of the series resemble the sequence of partial sums of a series whose nth term is 1/(3n-1). Thus, we can express the series as the sum of the terms of the form 1/(3n-1).

Next, we consider the integral ∫(1/(3x-1)) dx from 1 to infinity. To apply the integral test, we need to verify if this integral is convergent or divergent. We evaluate the integral using techniques of integration, which yields ln|3x-1|/3. Then, we take the limit of this expression as x approaches infinity and subtract the value of the expression as x approaches 1.

If the integral converges, then the series also converges. If the integral diverges, then the series also diverges. Therefore, by evaluating the integral, we can determine the convergence or divergence of the series.