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Tresset [83]
3 years ago
12

8. Can a series still converge if there is a boundary for n?

Mathematics
1 answer:
solong [7]3 years ago
8 0
\sin x> for 0, but \sin\dfrac\pi n is only decreasing for n\ge2, as you mentioned. This will not affect the behavior of the series. The sum starting from n=2 converges, so its value must be finite. Including the term generated by n=1 (which is finite, and zero besides, in this case) is the same as adding two finite numbers, which is yet another finite number. So the fact that the alternating series test only applies for a subset of the terms in the series doesn't matter.

For the second problem, the ratio test is an excellent choice. You have

\displaystyle\lim_{n\to\infty}\left|\dfrac{\frac{1\times3\times\cdots\times(2n-1)\times(2n+1)}{2\times5\times\cdots\times(3n-1)\times(3n+2)}}{\frac{1\times3\times\cdots\times(2n-1)}{2\times5\times\cdots\times(3n-1)}}\right|=\lim_{n\to\infty}\left|\frac{2n+1}{3n+2}\right|=\dfrac23

which means the series converges.
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Let y represent the number of adults that used the public pool that day.

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1.50 for kids and $2.25 for adults. The receipt totaled to $390. This means that

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Substituting x = 214 - y into equation 1, it becomes

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