How do you find the positive values of p for which #Sigma (1/n(lnn)^p)# from #[2,oo)# converges?

Answer 1

The series:

#sum_(n=2)^oo (lnn)^p/n#

is convergent for #p <= -1#

Consider the function:

#f(x) = (lnx)^p/x > 0# for #x in [2,oo)#
For #p < 0# we have:
#lim_(x->oo) (lnx)^p/x = lim_(x->oo) 1/(x(lnx)^absp) = 0#
While for #p > 0# the limit:
#lim_(x->oo) (lnx)^p/x#
is in the indeterminate form #oo/oo# so we can solve it using l'Hospital's rule:
#(1) lim_(x->oo) (lnx)^p/x = lim_(x->oo) (d/dx(lnx)^p)/(d/dx x) = lim_(x->oo) (p(lnx)^(p-1))/x#
For #p in NN# we can then prove by induction that:
#lim_(x->oo) (lnx)^p/x = 0#
since this is true for #p=0#, and #(1)# shows that if the limit is zero for #(p-1)# then it is zero also for #p#.
For any other #p > 0# we can see that if # m < p < n# with #m,n in NN^0#, we have either that:
# (lnx)^m/x < (lnx)^p/x < (lnx)^n/x#

or that:

# (lnx)^n/x < (lnx)^p/x < (lnx)^m/x#

In both cases by the squeeze theorem:

#lim_(x->oo) (lnx)^p/x = 0#
which then is true for any #p in RR#

Now consider the derivative:

#(df)/dx = (p(lnx)^(p-1)-(lnx)^p)/x^2 = (lnx)^(p-1)/x^2(p-lnx)#
we have that for #x > e^p#
#(df)/dx < 0#
so that the function is decreasing over the interval #(e^p,oo)#

All the hypotheses of the integral test are then satisfied and the convergence of the series:

#sum_(n=2)^oo (lnn)^p/n#

is equivalent to the convergence of the integral:

#int_2^oo (lnx)^p/x dx = int_2^oo (lnx)^p d(lnx) = [(lnx)^(p+1)/(p+1)]_2^oo#

Clearly the integral is convergent only if:

#lim_(x->oo) (lnx)^(p+1) < oo#
that is for #p <= -1#.
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Answer 2

To find the positive values of ( p ) for which the series ( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} ) converges, you can use the integral test:

  1. Integrate the function ( f(x) = \frac{1}{x(\ln x)^p} ) from 2 to infinity.
  2. Determine the convergence of the improper integral.

If the integral converges, then the series also converges. So, find the values of ( p ) for which the integral converges.

To integrate ( f(x) ), you may need to use substitution or other integration techniques. After integrating, analyze the convergence of the integral.

If the integral converges, then the series ( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} ) converges. Therefore, the values of ( p ) for which the integral converges will be the values for which the series converges.

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Answer 3

To determine the positive values of ( p ) for which the series ( \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} ) converges, you can use the integral test.

  1. Integrate the function ( \frac{1}{x(\ln x)^p} ) with respect to ( x ) from ( x = 2 ) to ( x = \infty ).
  2. Evaluate the integral.
  3. Determine for which values of ( p ) the integral converges.

If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Please note that finding the antiderivative of ( \frac{1}{x(\ln x)^p} ) and evaluating the integral may involve integration by parts or substitution, depending on the value of ( p ).

Once you've determined the integral and its convergence, you can find the values of ( p ) for which the series converges.

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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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