Question

How do you test the series \[\sum{\dfrac{\ln n}{n}}\] from n is \[[1,\infty )\] for convergence?

Answer 1

Hint: 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 \[\sum{f(n)}\] for the range \[[a,\infty )\] is divergent, here a is an integer. If the \[f(n)<1\] then the \[\sum{f(n)}\] for the range \[[a,\infty )\] 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 \[\ln x\] is strictly increasing in the range \[(e,\infty )\], here e is the exponential constant and \[e\approx 2.71\]. The nearest integer of e is 3, hence, we can say that for the range \[[3,\infty )\], \[\ln n>1\] n is an integer in the given range.
Dividing both sides of the inequality by n, we get
\[\Rightarrow \dfrac{\ln n}{n}>\dfrac{1}{n}\]
We know that the \[\sum{\dfrac{1}{n}}\] for \[[n,\infty )\] is a divergent series.
Thus, we can say that the \[\sum{\dfrac{\ln n}{n}}\] is also a divergent series for the given range.

Note:
Here one may ask that, we examine the series only for \[[3,\infty )\], 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.