- #1

RadiationX

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[tex]\sum_{n=1}^\infty\frac{(2n)!}{n^n}[/tex]

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- Thread starter RadiationX
- Start date

In summary, the conversation discusses the use of various convergence tests, such as the ratio test, for determining whether a series converges. The problems presented involve calculating the convergence of series involving factorials, cosines, and powers, and the conversation offers hints and guidance for solving them.

- #1

RadiationX

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[tex]\sum_{n=1}^\infty\frac{(2n)!}{n^n}[/tex]

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- #2

RadiationX

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[tex]\sum_{n=1}^\infty\frac{\cos{n\pi}}{n}[/tex]

[tex]\sum_{n=1}^\infty\frac{n^{1000}}{1.001^n}[/tex]

[tex]\sum_{n=1}^\infty\frac{n!}{1000^nn^{1000}}[/tex]

[tex]\sum_{n=2}^\infty(-1)^n\frac{n}{2n+1}[/tex]

all i need to know is if they converge. I don't need to find their sums, if they exist.

thanks in advance I'm at a loss as to what to do. these are due tomorrow afternoon.

- #3

RadiationX

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these must be some difficult problems! I guess i don't feel so bad now.

- #4

HallsofIvy

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- #5

RadiationX

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HallsofIvy said:

These are the test that i know.

1. nth term test for divergence

2. convergence of p-series

3. ratio test

4. integral test

5. root test

I have tried what i know. I guess I'm not seeing how to apply these test. I haven't posted here for quick answers because i want to know how to solve these problems myself. I just don't see how to solve them right now.

- #6

Justin Lazear

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Try the ratio test on the first problem and post how it goes.

--J

--J

- #7

RadiationX

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i'm not coming up with anything.

- #8

Justin Lazear

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Clearly, since you're asking the question. I meant post your work and the steps you followed.

--J

--J

- #9

RadiationX

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RadiationX said:

[tex]\sum_{n=1}^\infty\frac{(2n)!}{n^n}[/tex]

for this one i know that i should use the ratio test but i don't know how to simplify the factorial. I know this much

[tex]\frac{2(n+1)!}{(n+1)^{n+1}}\frac{n^n}{2n!}[/tex]

is this much correct?

- #10

RadiationX

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once i get past tomorrow i can devote about 12-14 hrs to getting caught up.

- #11

Justin Lazear

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You're looking for

[tex]\frac{(2(n+1))!}{(n+1)^{n+1}}\frac{n^n}{2n!}[/tex]

Note that (2n+2)! = (2n+2)(2n+1)(2n)! and that (n+1)^(n+1) = (n+1)(n+1)^n.

Can you simplify it from there?

--J

- #12

Justin Lazear

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For the second one, calculate [itex]\cos{n\pi}[/itex] for a few n. See a pattern? See if you can't represent the cosine as something more familiar, and your series should become familiar.

Third and fourth can both be determined by the ratio test.

For the fifth, consider the most basic conditions on [itex]a_n[/itex] for [itex]\sum{a_n}[/itex] to converge.

--J

- #13

RadiationX

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Justin Lazear said:

For the second one, calculate [itex]\cos{n\pi}[/itex] for a few n. See a pattern? See if you can't represent the cosine as something more familiar, and your series should become familiar.

Third and fourth can both be determined by the ratio test.

For the fifth, consider the most basic conditions on [itex]a_n[/itex] for [itex]\sum{a_n}[/itex] to converge.

--J

the third one is inconclusive using the ratio test because the limit of the ratio is equal to 1

- #14

torchbear

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RadiationX said:the third one is inconclusive using the ratio test because the limit of the ratio is equal to 1

That is not true. The ratio tends to be infinite (it is compatitable with n).

- #15

Justin Lazear

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torchbear said:

That is not true. The ratio tends to be infinite (it is compatitable with n).

The third one converges and the limit of the ratio is less than 1, it is 1/1.001.

--J

The Ratio Test is a method used to determine whether an infinite series converges or diverges. It compares the ratio of consecutive terms in a series to a limiting value, and if this limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and another method must be used.

To apply the Ratio Test, you must first find the ratio of consecutive terms in the series. Then, take the limit of this ratio as n approaches infinity. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive and another method must be used.

The limiting value in the Ratio Test is the critical value that determines whether a series converges or diverges. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive and another method must be used.

The Ratio Test can be applied to any series with positive terms, including power series, geometric series, and trigonometric series. However, it cannot be used on alternating series or series with negative terms.

Yes, the Ratio Test can be used to prove absolute convergence. If the limit of the ratio of consecutive terms is less than 1, the series converges absolutely. This means that the series converges regardless of the sign of its terms.

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