Question

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

Answer 1

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.