Does xn Converge? A Comparison Test

In summary, the conversation discusses the convergence of the series xn = 1/(n + SQRTn) using the comparison test. It is determined that the series diverges since the power of n is less than one. There is some confusion about whether to use a non-zero or zero constant, but it is ultimately clarified that a non-zero constant will still result in divergence.
  • #1
Mattofix
138
0

Homework Statement



Does xn converge (Sum from n=1 to infinity) of xn = 1/(n + SQRTn)

Homework Equations



Using comparision test

The Attempt at a Solution



I separted into fractions of 1/SQRTn - 1/(1 + SQRTn) and i know that both of these diverge since the power of n is less than one but am stuck as to whether is converges or diverges and how to prove it...
 
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  • #2
[tex]\frac{1}{n+n}\leq \frac{1}{n+\sqrt{n}}[/tex]
 
  • #3
therefore xn divereges...?
 
  • #4
yeah that is right, since the harmonic series diverges, it also diverges when we multiply it by a constant.
 
  • #5
sutupidmath said:
yeah that is right, since the harmonic series diverges, it also diverges when we multiply it by a constant.
Non-zero constant.

Not to offend you but that is what I like to call a "physics-type mistake".
 
  • #6
Kummer said:
Non-zero constant.

Not to offend you but that is what I like to call a "physics-type mistake".

yeah that is what i actually meant, but thnx for pointing it out. and not am not offended in any way.
 

1. What is the comparison test for determining convergence of a series?

The comparison test is a method used to determine the convergence or divergence of a series by comparing it to another series whose convergence or divergence is already known. It states that if the terms of the series in question are always less than or equal to the terms of a convergent series, then the original series must also converge. Similarly, if the terms are always greater than or equal to the terms of a divergent series, then the original series must also diverge.

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

The direct comparison test compares the terms of the series in question directly to the terms of a known convergent or divergent series, while the limit comparison test compares the ratio of the terms of the two series as the index approaches infinity. The direct comparison test is typically easier to use, but the limit comparison test can be useful when the terms of the series do not have a clear comparison.

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

No, the comparison test only tells us whether a series converges or diverges. To determine the exact value of a convergent series, other methods such as the geometric series test or the integral test must be used.

4. Are there any limitations to using the comparison test?

Yes, the comparison test can only be used when the terms of the series are always positive. It also cannot be used to determine the convergence or divergence of alternating series.

5. Can the comparison test be used for infinite series with negative terms?

No, the comparison test can only be used for series with positive terms. However, the absolute convergence test can be used to determine the convergence of series with both positive and negative terms.

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