Question

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.

Explanation:

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:

• `f(n)` must be decreasing on `n in [N,oo)`
• `f(n)>0` for `n in [N,oo)`

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.