Question
Question: How do you test the series \[\sum\limits_{2}^{\infty }{\dfrac{1}{\ln \left( n! \right)}}\] where n i...
How do you test the series 2∑∞ln(n!)1 where n is from [2,∞] for convergence?
Solution
To solve the question that is stated above we will use the d’Alembert’s ratio test where we know the equation and we will use limits where n tends to ∞ for the mentioning equation and if the value of the limit is less than 1 there is convergence taking place, if it is more than 1 divergence is taking place, if it is equal to 1 than the limit fails to exist and the test is inconclusive.
Complete step-by-step solution:
So as we begin with the d’Alembert’s ratio test we will apply the limits to the function provided and from this we get: n→∞limln(n!)1
In this limit when we substitute the value of n which is been stated as tending towards ∞ we can clearly see that the limits is also tending towards ∞ as we know that factorial of any number increases as the number of the factorial increases i.e. the value of 4! is greater than 3! And the difference between the values of both also has a huge difference so when we take the value to be ∞ we can clearly say that the value also tends towards infinity and now we can say that the value inside the logarithmic function is ∞, now to find the value of ln(n!) which is basically ln(∞), as we know that the value infinity is very huge and has no significant numeric form so the end result of the logarithmic value will also become ∞and we all know ∞1 tends towards 0, which gives us the value of the limit as less than 1 and hence there is convergence taking place.
So, as the value of limits is tending towards 0 and hence is less than 1 we can say that there is convergence taking place from the d’Alembert’s ratio test.
Note: The common mistakes that will happen in the question is while finding the limits of the function provided, we got it to come as 0, but if you have a function and the answer comes to be ∞ or 00 use the L’hospital’s rule to find the limits.