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How do you use the integral test to determine if #Sigma lnn/n^2# from #[1,oo)# is convergent or divergent?

Answer 1

William Abdalla

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

If we use:

#f(x) = lnx/x^2#

as test function, we have that for #x in [1,+oo)#:

# f(x) # is positive

# lim_(x->oo) f(x) = 0 #

#f(n) = lnn/n^2#

Calculating the first derivative:

#f'(x) = frac (x^2*1/x - 2xlnx) (x^4) = (1-2lnx)/x^3#

we can also see that #f(x)# is strictly decreasing for #x>sqrt(e)#

so all the hypotheses of the integral test theorem are satisfied and the series is convergent if the integral:

#int_1^oo lnx/x^2dx#

also converges.

Let's calculate the indefinite integral by parts:

#int lnx/x^2dx = int lnxd(-1/x) = -lnx/x + int 1/xd(lnx) = -lnx/x + int dx/x^2 = -lnx/x - 1/x +C = -(lnx+1)/x + C #

So:

#int_1^oo lnx/x^2dx = [-(lnx+1)/x] _1^oo = 0 - (-1) = 1#

That is the integral is convergent and so is the series.

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