Question

How do you use the limit comparison test to determine if \[\sum{\tan \left( \dfrac{1}{n} \right)}\] from \[[1,\infty )\] is convergent or divergent?

Answer 1

Hint: In the above question we have to check whether the series is convergent or divergent. By applying the basic definition of limit comparison test we can find the nature of the series. Also, in the given function we can see that when n approaches infinity the value of the function will be approximately equal to \[\dfrac{1}{n}\] and this is a divergent function.

Complete step by step answer:
The above question belongs to the concept of calculus. Here we have to use a limit conversion test to check the nature of the given series. The limit conversion test states that suppose if we have a series then the value of limit to the ratio of the general term of the series and inverse of its argument value. If the value is some finite number greater than zero, then either both series converges or both diverge. if the value is equal to zero then the series is convergent and if the value is one then the series is divergent.
 Now in the question, the series given is \[\sum{\tan \left( \dfrac{1}{n} \right)}\]
Here \[{{a}_{n}}=\tan \left( \dfrac{1}{n} \right),{{b}_{n}}=\dfrac{1}{n}\]
Now applying a limit conversion test.
\[\displaystyle \lim_{n \to \infty }\dfrac{{{a}_{n}}}{{{b}_{n}}}=\displaystyle \lim_{n \to \infty }\dfrac{\tan \left( \dfrac{1}{n} \right)}{\dfrac{1}{n}}=1\]
As the value is equal to one
Therefore, the given series is divergent as per the limit conversion test.

Note:
While solving the above question keep in mind the condition for a series to be divergent or convergent. Also, keep in mind the nature of \[\tan x\] function. As for small values of x, \[\tan x\] gives a value approximately equal to the argument which is x. try to remember the limit conversion test for future use. Also, \[{{a}_{n}}\] and \[{{b}_{n}}\] should be positive and greater than zero.