How do you test the series #Sigma lnn/(nsqrtn)# from n is #[1,oo)# for convergence?

Answer 1

The series:

#sum_(n=1)^oo lnn/(nsqrtn)#

is convergent.

You can use the integral test, choosing as test function:

#f(x) = lnx/(xsqrtx)#

We can verify that:

(i) #f(x) > 0# for #x in (1,+oo)#
(ii) #lim_(x->oo) lnx/(xsqrtx) = 0#
The limit is in the indeterminate form #oo/oo# so we can use l'Hospital's rule to solve it:
#lim_(x->oo) lnx/(xsqrtx) = lim_(x->oo) (d/(dx) lnx )/(d/(dx) x^(3/2)) = lim_(x->oo) (1/x) 1/(3/2x^(1/2)) = lim_(x->oo) 2/(3x^(3/2)) = 0#
(iii) #f(x)# is monotone decreasing for #x in (1+oo)#

We can calculate the first derivative:

#d/(dx) lnx/(xsqrtx) = d/(dx) lnx *x^(-3/2) = (1/x)x^(-3/2) -3/2x^(-5/2)lnx = x^(-5/2) (1-3/2lnx) = (2-3lnx)/(2x^(5/2))#
and we can see that #x > 1 => f'(x) < 0#
(iv) #f(n) = lnn/(nsqrt(n))#

Thus all the hypotheses of the integral test theorem are satisfied and the convergence of the series is equivalent to the convergence of the improper integral:

#int_1^oo lnx/(xsqrtx)dx#

We solve the indefinite integral by parts:

#int lnx * x^(-3/2) dx = int lnx d(-2x^(-1/2)) =-2lnx/sqrt(x) +2 int x^(-1/2)/xdx = -2lnx/sqrt(x) +2int x^(-3/2)dx = -2lnx/sqrt(x) -4 x^(-1/2) +C = - (2(lnx+2))/sqrt(x)+C#

and we have:

#int_1^oo lnx/(xsqrtx)dx = [- (2(lnx+2))/sqrt(x)]_1^oo #
#int_1^oo lnx/(xsqrtx)dx = 4 - lim_(x->oo) lnx/(xsqrtx)#

We have already calculated at point (ii) above that such limit is zero, then:

#int_1^oo lnx/(xsqrtx)dx = 4#

proving that the series is convergent.

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Answer 2

You can test the series ( \sum \frac{\ln(n)}{n\sqrt{n}} ) for convergence using the Limit Comparison Test or the Ratio Test.

Using the Limit Comparison Test:

  1. Choose a known convergent series ( \sum b_n ) such that ( b_n ) is positive for all ( n ) and ( \lim_{n \to \infty} \frac{a_n}{b_n} ) exists and is positive, where ( a_n = \frac{\ln(n)}{n\sqrt{n}} ).
  2. Compute ( \lim_{n \to \infty} \frac{a_n}{b_n} ).
  3. If the limit is finite and positive, both series either converge or diverge. If it's zero or infinite, the comparison test is inconclusive.

Using the Ratio Test:

  1. Compute ( \lim_{n \to \infty} \frac{a_{n+1}}{a_n} ), where ( a_n = \frac{\ln(n)}{n\sqrt{n}} ).
  2. If the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit does not exist, the series diverges. If the limit equals 1, the test is inconclusive.

After applying either of these tests, you can determine the convergence or divergence of the series ( \sum \frac{\ln(n)}{n\sqrt{n}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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