Question
Question: How do you test the series \[\sum{\dfrac{\ln n}{n}}\] from n is \[[1,\infty )\] for convergence?...
How do you test the series ∑nlnn from n is [1,∞) for convergence?
Solution
To find whether a given expression is convergent or not. We need to check if the given f(n) is greater than one or less than one. If f(n)>1 then the ∑f(n) for the range [a,∞) is divergent, here a is an integer. If the f(n)<1 then the ∑f(n) for the range [a,∞) is convergent, here a is an integer. Thus, we first need to find in which range the expression does the function in the summation lies.
Complete step by step solution:
We know that the logarithmic function lnx is strictly increasing in the range (e,∞), here e is the exponential constant and e≈2.71. The nearest integer of e is 3, hence, we can say that for the range [3,∞), lnn>1 n is an integer in the given range.
Dividing both sides of the inequality by n, we get
⇒nlnn>n1
We know that the ∑n1 for [n,∞) is a divergent series.
Thus, we can say that the ∑nlnn is also a divergent series for the given range.
Note:
Here one may ask that, we examine the series only for [3,∞), then what about the range [1,3]. We should note that this part will not affect the final result as this range is very small compared to the range we take. And even if this part gives a convergent result, we will use the property that the series which is the summation of convergent and divergent series is divergent.
Hence, we can do this.