The infinite series,
Is slightly famous.
(Called Bertrand Series)
The results says that if,
---> converges
---> diverges
---> depends.
The value of makes the dependance.
If then it converges only if .
Therefore, your series converges when
The function,
Is a positive, decreasing, continous function for
With the property that,
in the series.
You can apply the integral series test.
And see when, (I am just goint to change the lower limit, it makes no difference for convergence).
If,
then,
Thus, by the substitution rule,
Since the antiderivative depends whether is 1 or not.
So we will divide this problem into 3 cases,
It, then,
Thus, (dropping the last term, makes no difference)
Which will increase without bound for,
If then,
Thus, (dropping the last term, makes no difference)
.
Which will increase without bound for,
is increasing (though very very very, snail slowly).
If then,
Thus, (dropping the last term, makes no difference)
Which will converge for, and this assures us the fraction is in the denominator with the denominator increasing without bound, so the limit converges.
The the solution is based on,
First for and (note the reason why I make these inequalities is because otherwise the function might not be defined or I will not be able to get integrals).
Then,
Thus, the integral becomes,
Express this integral as,
If, then
Thus,
Substitution rule,
Express as,
Let,
then, thus,
By the substitution rule we have,
Thus,
Thus,
is the integral.
Thus, the original integral is,
How! Did you try to solve it? No.
Look at the integral. For the same reason it will diverge for for is divergent.
And if the exponent then you have again an increasing function so it is divergent. When then the exponent is negative thus the function will eventually die. In that case is convergese for