Finding divergence/convergence by direct comparison test

In summary, the conversation discusses the comparison test and how to use it to show that \sum^{∞}_{1}1/n^{n} converges. The comparison is made to the p series \sum^{∞}_{1}1/n^{2} and the need to show that \sum^{∞}_{1}1/n^{n} is smaller than \sum^{∞}_{1}1/n^{2} by using the comparison test. The conversation also mentions finding nice factors to compare the equation to 1.
  • #1
OnceKnown
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Homework Statement


[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex]


Homework Equations

Direct comparison test



The Attempt at a Solution

Since the main factor in the equation is the exponent that would be changing as n goes to infinity, I know that from the p series as p > 1 the the series converges. So I know that I would be comparing the original equation to

[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

And I know that I need to show:

0 [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

but I don't know how to show

[itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

mathematically. Would I just blatantly say that the original term is smaller than the p series just by looking at it?
 
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  • #2
OnceKnown said:
And I know that I need to show:

0 [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

The comparison test requires you to show that

[tex] \left| \frac{1}{n^n} \right| \leq \left| \frac{1}{n^2} \right| [/tex]

for sufficiently large [itex]n[/itex]. Try to find nice factors that you can multiply both sides of the equation to compare something to 1.
 

What is the direct comparison test and how does it work?

The direct comparison test is a method used to determine whether a given series is convergent or divergent by comparing it to a known convergent or divergent series. This is done by finding a series that is larger than the given series and another series that is smaller than the given series, and then comparing their convergence or divergence. If the larger series is convergent, then the given series must also be convergent. If the smaller series is divergent, then the given series must also be divergent.

When should the direct comparison test be used?

The direct comparison test should be used when the terms of a series can be easily compared to the terms of a known convergent or divergent series. It is also useful when the given series can be rewritten in a way that makes it easier to compare to a known series. If the terms of the given series are difficult to compare or if the series does not resemble any known series, then other convergence or divergence tests should be used.

What is the difference between the direct comparison test and the limit comparison test?

The direct comparison test compares the given series to a known series directly, while the limit comparison test compares the ratio of the terms of the given series to the terms of a known series. The direct comparison test is generally easier to use, but the limit comparison test can be applied in more cases.

Can the direct comparison test be used to determine the exact value of a convergent series?

No, the direct comparison test can only be used to determine whether a series is convergent or divergent. It cannot be used to find the exact value of a convergent series.

What are some common mistakes when using the direct comparison test?

One common mistake is comparing the given series to a known series that is not the correct type of series (i.e. comparing an alternating series to a non-alternating series). Another mistake is using a known series that is not strictly larger or smaller than the given series. It is important to carefully choose the known series for an accurate comparison.

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