(After all, to show a statement is false we only need one counterexample. It does not have to be all $\varepsilon >0$. Use the fact that lim n ( n n + k) n e n. We only need to find one such $\varepsilon $. Convergent Sequence.Convergent Sequence in real analysis.Convergent sequence in hindi.Convergent Sequence definition.Convergent Sequence examples.Convergent. Confirm that the series, n 1 n n n, is absolutely convergent. Now, if we can find an $\varepsilon $ so that this last implication is false we get a contradiction - which is what we seek. This is because we are assuming that the sequence is convergent and hence for every $\varepsilon >0$ we can find some $N\in \N$ so that $n>N\Longrightarrow |a_n-a|\lt \varepsilon$. In our previous examples we had to deal with all $\varepsilon $ greater than zero - we had no choice in the matter. What is confusing for many students is that we chose $\varepsilon$ in the example. As a result, the sum, difference, product and ratio of two convergent sequences automatically converge (if were not dividing by numbers close to zero), as do. Hence, $(-1)^n$ has no limit as $n\to \infty $. This is a contradiction since $2\lt 2$ is definitely false. Then, by the triangle inequality we have, A sequence converges when it keeps getting closer and closer to a certain value. Then, there exists $N$ such that $|(-1)^n -a|\lt 1$ for all $n>N$. For example, the sequence (2, 2, 2, ) converges to 2. Let $a$ be the limit of $(-1)^n$ as $n\to \infty $. All constant sequences converge to the real-numbered term in the sequence. To show that $a_n$ does not have a limit we shall assume, for a contradiction, that it does. The sequence $a_n=(-1)^n$ is not convergent. Let’s now see some examples of sequences that do not converge, i.e., they are non-convergent sequences.
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