For $x_n=\frac{(-1)^n}{n}$, my professor said that the inf for when n is even is $\frac{-1}{n+1}$ and the sup is $\frac{1}{n}$. How is the inf a negative number since $(-1)^n$ where n is an even number is always positive?
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What your professor said, if that is exactly what he said, makes zero sense. The infimum and supremum are fixed number and should not include $n$. If you take $n$ to be even, as you say, you always have positive terms. The supremum is $\frac{1}{2}$ which is for $n=2$. You can see this, since for all larger $n$, the sequence becomes smaller. The infimum would be $0$. Now, for the odd case, you always have negative numbers, so the infimum would be $-1$ and the supremum $0$.
The whole idea behind the supremum and the infimum of a sequence, is that you consider all the terms of the sequence at once, and want to find the largest and smallest one. However, sometimes, there is no largest or smallest element, so the supremum or infimum do not actually belong in the set. Here, you can see this since $0$ is not an actual element of the sequence.
Esoog
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So I do still need an inf and sup for even and odd numbers? – ematth7 Oct 12 '15 at 01:56
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Can the inf and sup ever be infinity? For example, if S={$2*(-1)^n+(\frac{-1}{n})^{n+1}$, would the inf for when n is odd=-infinity and the sup for when n is even equal infinity? – ematth7 Oct 12 '15 at 02:16
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I don't understand completely the example you gave, but from what it looks like, no infimum or supremum are infinity since you have alternating 2, -2 and the other part goes to 0. However, you can definitely have infinity as supremum or infimum. Just take any unbounded sequence, for example the sequence n, which is just 1,2,3,4,.... Or you can have $(-1)^n *n$ which alternates and thus supremum is infinite, and infimum would be minus infinite. – Esoog Oct 12 '15 at 02:27
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Your teacher is flatly wrong ! the $\inf(x_n) = 0$ when $n$ is even, and $\sup(x_n) = \dfrac{1}{2}$ when $n$ is also even
DeepSea
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That's the answer that I put and he marked it wrong and then told me the above. – ematth7 Oct 12 '15 at 01:54