How do you use the integral test to determine if #1/(sqrt1(sqrt1+1))+1/(sqrt2(sqrt2+1))+1/(sqrt3(sqrt3+1))+...1/(sqrtn(sqrtn+1))+...# is convergent or divergent?

Answer 1

The series is divergent. See the explanation below.

This is the series:

#sum_(n=1)^oo1/(sqrtn(sqrtn+1))#

Before starting with the integral test, we need to see that a couple conditions are met first.

In order for #sum_(n=N)^oof(n)# to be testable via the integral test, we see that:
For #f(n)=1/(sqrtn(sqrtn+1))# we see that #f(n)>0# on #n in [1,oo)# because there are never any negative terms. Furthermore, as #n# increases, the denominator increases, meaning that #f(n)# as a whole is always decreasing on the interval #n in [N,oo)#. So, the conditions are met for the integral test to apply.
The integral states that if the improper integral #int_N^oof(x)dx# converges (which, for an integral, means that the integral, when evaluated, equals any finite value), then the series converges as well. If the integral diverges, then the series does as well.
So, in order to test the convergence of the given series, we need to evaluate the integral #int_1^oo1/(sqrtx(sqrtx+1))dx#. Let's first find the integral without the bounds.
#int1/(sqrtx(sqrtx+1))dx" "" "#Let: #{(u=sqrtx+1),(du=1/(2sqrtx)dx):}#
#=2int1/(2sqrtx(sqrtx+1))dx=2int1/udu=2lnabsu+C#

Then:

#int_1^oo1/(sqrtx(sqrtx+1))dx" "" "" "#(Take the limit at infinity.)
#=lim_(brarroo)int_1^b1/(sqrtx(sqrtx+1))dx=lim_(brarroo)(2lnabsu)|_1^b#
#=lim_(brarroo)2lnb-2ln1#
As #brarroo#, we see that #lnbrarroo# as well:
#=oo#
The integral diverges. Through the integral test, the series #sum_(n=1)^oo1/(sqrtn(sqrtn+1))# diverges as well.
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Note that #sum_(n=1)^oo1/(sqrtn(sqrtn+1))=sum_(n=1)^oo1/(n+sqrtn)#
This can be alternatively be compared via the limit comparison test with #sum_(n=1)^oo1/n#, a divergent series, to prove the divergence of the given series as well.

This is just as valid a method as the integral test, but quicker.

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Answer 2

To use the integral test to determine the convergence or divergence of the series 11(1+1)+12(2+1)+13(3+1)++1n(n+1)+\frac{1}{\sqrt{1}(\sqrt{1}+1)} + \frac{1}{\sqrt{2}(\sqrt{2}+1)} + \frac{1}{\sqrt{3}(\sqrt{3}+1)} + \ldots + \frac{1}{\sqrt{n}(\sqrt{n}+1)} + \ldots:

  1. Check if the series terms are positive, which they are.
  2. Verify that the series terms are decreasing, which they are.
  3. Integrate the function f(x)=1x(x+1)f(x) = \frac{1}{\sqrt{x}(\sqrt{x} + 1)} from x=1x = 1 to x=x = \infty.
  4. Determine if the integral converges or diverges.

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

By integrating the function f(x)f(x) and evaluating the integral, you can determine the convergence or divergence of the given series.

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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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