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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 lnnn\sum{\dfrac{\ln n}{n}} from n is [1,)[1,\infty ) for convergence?

Explanation

Solution

To find whether a given expression is convergent or not. We need to check if the given f(n)f(n) is greater than one or less than one. If f(n)>1f(n)>1 then the f(n)\sum{f(n)} for the range [a,)[a,\infty ) is divergent, here a is an integer. If the f(n)<1f(n)<1 then the f(n)\sum{f(n)} for the range [a,)[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 lnx\ln x is strictly increasing in the range (e,)(e,\infty ), here e is the exponential constant and e2.71e\approx 2.71. The nearest integer of e is 3, hence, we can say that for the range [3,)[3,\infty ), lnn>1\ln n>1 n is an integer in the given range.
Dividing both sides of the inequality by n, we get
lnnn>1n\Rightarrow \dfrac{\ln n}{n}>\dfrac{1}{n}
We know that the 1n\sum{\dfrac{1}{n}} for [n,)[n,\infty ) is a divergent series.
Thus, we can say that the lnnn\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,)[3,\infty ), then what about the range [1,3][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.