Question

How do you determine whether the infinite sequence ${{a}_{n}}=n.\cos \left( n\text{ }\text{. }\pi \right)$ converges or diverges.

Answer 1

In this problem, we have to find whether the given infinite sequence converges or diverges. We know that an infinite sequence of numbers is an ordered list of numbers with an infinite number of numbers. We also know that the sequence \left\\{ {{a}_{n}} \right\\} converges if $\displaystyle \lim_{n \to \infty }{{a}_{n}}$ exists (having a finite value) otherwise it diverges. We can first apply a limit for the subsequence of the given sequence, to check whether it converges or diverges.

Complete step-by-step solution:
We know that the given infinite sequence is,
${{a}_{n}}=n.\cos \left( n\text{ }\text{. }\pi \right)$
We also know that the sequence \left\\{ {{a}_{n}} \right\\} converges if $\displaystyle \lim_{n \to \infty }{{a}_{n}}$ exists (having a finite value) otherwise it diverges when $\displaystyle \lim_{n \to \infty }{{a}_{n}}$ has infinite values.
We can now write the subsequence of the above infinite sequence of only even terms, we get
${{a}_{2n}}=2n.\cos \left( 2n\text{ }\text{. }\pi \right)$
We can now take limit on both sides of its subsequence of only even terms, we get
$\Rightarrow \displaystyle \lim_{n \to \infty }{{a}_{2n}}=\displaystyle \lim_{n \to \infty }\left[ 2n.\cos \left( 2n\text{ }\text{. }\pi \right) \right]$
We know that $\cos \left( 2n\pi \right)=1$ for all integers n.
$\Rightarrow \displaystyle \lim_{n \to \infty }\left( 2n \right)=1\times 2\left( \infty \right)=\infty$
We can see that the above $\displaystyle \lim_{n \to \infty }{{a}_{2n}}$ has infinite value and it diverges.
We should know that since subsequence diverges, the original sequence also diverges.
Therefore, the given sequence ${{a}_{n}}=n.\cos \left( n\text{ }\text{. }\pi \right)$ diverges.

Note: Students make mistakes while substituting the limit values and finding the final answer. We should always remember that the sequence \left\\{ {{a}_{n}} \right\\} converges if $\displaystyle \lim_{n \to \infty }{{a}_{n}}$ exists (having a finite value) otherwise it diverges when $\displaystyle \lim_{n \to \infty }{{a}_{n}}$ has infinite values. We should also remember that when the subsequence diverges, the original sequence also diverges.