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Analyze the convergence or divergence of the following sequence

a) $\left\{\frac{1}{n}+\sin\frac{n\pi}{2}\right\}$

The first one is divergent because of the in $\sin\frac{n\pi}{2}$ term, which takes the values, for $n = 1, 2, 3, 4, 5, \dots$:

$$1, 0, -1, 0, 1, 0, -1, 0, 1, \dots$$

As you can see, it's divergent. To formally prove it,I could simply notice that it has constant subsequences of $1$s, $0$s, and $-1$s, all of which converge to different limits. If it were as subsequence, they would all be the same limit.

My procedure is that correct?

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Is it really so hard to remember that the post starts in the body, not the title? Given that you are getting your homework pre-checked for free, perhaps you might be so kind as to show some consideration to the readers. – Arturo Magidin May 24 '12 at 12:03
up vote 3 down vote accepted

You could also use the fact that if a sequence is convergent, then the difference between successive terms must converge to $0$.

Let $x_n = \frac{1}{n}+\sin \frac{n \pi}{2}$. Then $x_{n+1}-x_n = \cos \frac{n \pi}{2} - \sin \frac{n \pi}{2} -\frac{1}{n(n+1)}$, from which we get the estimate $|x_{n+1}-x_n| \geq |\cos \frac{n \pi}{2} - \sin \frac{n \pi}{2}| -\frac{1}{n(n+1)} = 1-\frac{1}{n(n+1)}$. So the difference does not converge to $0$, so you can conclude that the sequence is divergent.

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You’re on the right track, but you’ve left out an important step: you haven’t said anything to take the $1/n$ term into account. It’s obvious what’s happening, but you still have to say something.

Let $a_n=\frac1n+\sin\frac{n\pi}2$. If $\langle a_n:n\in\Bbb Z^+\rangle$ converged, say to $L$, then the sequence $\left\langle a_n-\frac1n:n\in\Bbb Z^+\right\rangle$ would converge to $L-0=L$, because $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ converges to $0$. Now make your (correct) argument about $\left\langle\sin\frac{n\pi}2:n\in\Bbb Z^+\right\rangle$ not converging and thereby get a contradiction. Then you can conclude that $\langle a_n:n\in\Bbb Z^+\rangle$ does not converge.

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Yes you are right for a convergent sequence all subsequences converge to the same value!

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Yes you are correct as all subsequences of convergent sequence of real numbers converge to the same limit.

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