How do you use the integral test to find whether the following series converges or diverges #sum( 1/(n*ln(n)^0.5) )#?

Answer 1

Does not converge

The function # 1/(x*sqrt( ln(x))# is positive, continuous, monotone decreasing over the interval #[N to oo)#, #N in mathbf Z^+# and so # sum _{n=N}^{oo } \ 1/(n*sqrt( ln(n)) # converges to a real number iff the following improper integral is finite:
#int_N^(oo) 1/(x*sqrt( ln(x)) \ dx#
#= 2 int_N^(oo) d(sqrt (ln(x))) #
#= 2 [ \ sqrt (ln(x)) \ ]_N^(oo)#
#= 2 ( sqrt (ln(omega)) |_(omega to oo) - sqrt (ln(N)\ ) \ )#
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Answer 2

To use the integral test to determine the convergence or divergence of the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ), we compare it to the integral of the corresponding continuous function.

First, we note that the function ( f(x) = \frac{1}{x \sqrt{\ln(x)}} ) is continuous, positive, and decreasing for ( x \geq 2 ), which corresponds to the terms of the series starting from ( n = 2 ).

Next, we consider the integral ( \int_2^\infty \frac{1}{x \sqrt{\ln(x)}} , dx ). To evaluate this integral, we can make a substitution ( u = \ln(x) ), which gives ( du = \frac{1}{x} , dx ). The integral becomes:

[ \int_{\ln(2)}^\infty \frac{1}{\sqrt{u}} , du ]

This integral converges because it is the integral of ( \frac{1}{\sqrt{u}} ), which is a p-series with ( p = \frac{1}{2} ) and ( p > 0 ).

Since the integral converges, by the integral test, the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ) also converges.

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Answer 3

To use the integral test to determine the convergence or divergence of the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ), we need to:

  1. Check if the function ( f(x) = \frac{1}{x \sqrt{\ln(x)}} ) is continuous, positive, and decreasing for ( x \geq 2 ).
  2. Evaluate the integral ( \int_{2}^{\infty} \frac{1}{x \sqrt{\ln(x)}} , dx ).
  3. Determine if the integral converges or diverges.

If the integral converges, then the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ) converges. If the integral diverges, then the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ) also diverges.

In this case, the function ( f(x) = \frac{1}{x \sqrt{\ln(x)}} ) is continuous, positive, and decreasing for ( x \geq 2 ).

Next, we evaluate the integral ( \int_{2}^{\infty} \frac{1}{x \sqrt{\ln(x)}} , dx ). This integral can be challenging to compute analytically, but we can use a comparison test to determine its convergence or divergence.

Since ( \frac{1}{x \sqrt{\ln(x)}} < \frac{1}{x} ) for ( x \geq 2 ), and the integral ( \int_{2}^{\infty} \frac{1}{x} , dx ) diverges (harmonic series), by the comparison test, the original integral ( \int_{2}^{\infty} \frac{1}{x \sqrt{\ln(x)}} , dx ) also diverges.

Therefore, by the integral test, the series ( \sum \frac{1}{n \sqrt{\ln(n)}} ) also diverges.

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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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