Why does 1/[nlog(n+1)] diverge

  • Thread starter grossgermany
  • Start date
In summary, the series 1/[nlog(n+1)] diverges and does not lend itself to being solved by the integral test. Additionally, it can be compared to the series 1/(n*log(n)) shifted one term, which can be used in a comparison test to show that the series diverges.
  • #1
grossgermany
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Homework Statement



Why does the series 1/[nlog(n+1)] diverge

Homework Equations


We know that 1/[nlog(n)] diverges by the integral test. However the question as written does not lend itself to be any integral precisely.


The Attempt at a Solution

 
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  • #2
investigate the integral:

[tex]
f(x) = \int_{1}^{x}{\frac{dt}{t \, \log{t}}}
[/tex]

as [itex]x \rightarrow \infty[/itex].
 
  • #3
Yes, however the integral doesn't work in this particular case
Note that the question says nlog(n+1), not nlogn
 
  • #4
grossgermany said:
Yes, however the integral doesn't work in this particular case
Note that the question says nlog(n+1), not nlogn

1/((n+1)*log(n+1)) diverges, right? You can do an integral test or just notice it's the same series as 1/(n*log(n)) shifted one term. Now try and think of how to use a comparison test.
 

Related to Why does 1/[nlog(n+1)] diverge

Why does 1/[nlog(n+1)] diverge?

There are several reasons why 1/[nlog(n+1)] may diverge. One possible reason is that the logarithm function, log(n+1), increases very slowly as n increases. This means that the denominator, nlog(n+1), will approach infinity as n approaches infinity, causing the overall fraction to approach zero. Another reason is that the log(n+1) term does not have a limit as n approaches infinity, which can also lead to divergence.

Can you explain the concept of divergence?

Divergence refers to the behavior of a sequence or series as its terms increase or decrease. If the terms of a sequence or series become infinitely large, the sequence or series is said to diverge. This means that it does not have a finite limit or sum and its behavior cannot be predicted.

What is the relationship between the divergence of 1/[nlog(n+1)] and the divergence of 1/n?

There is a close relationship between the divergence of 1/[nlog(n+1)] and the divergence of 1/n. In fact, 1/[nlog(n+1)] can be seen as a more slowly diverging version of 1/n. As n approaches infinity, the log(n+1) term in the denominator of 1/[nlog(n+1)] will cause the overall fraction to approach zero at a slower rate compared to 1/n, which will approach infinity. Therefore, 1/[nlog(n+1)] diverges, but at a slower rate than 1/n.

Are there any other functions that can cause 1/[nlog(n+1)] to diverge?

Yes, there are other functions that can cause 1/[nlog(n+1)] to diverge. For example, if the denominator contains a higher power of n, such as n^2 or n^3, the fraction will diverge at a faster rate. Additionally, if the numerator contains a function that increases faster than n, such as n^2 or 2^n, the fraction will also diverge at a faster rate.

How does the divergence of 1/[nlog(n+1)] compare to other commonly studied divergent series?

1/[nlog(n+1)] is often considered a relatively slow-diverging series compared to other commonly studied divergent series, such as the harmonic series (1/n) or the geometric series (1/2^n). However, it is faster diverging than some other commonly studied series, such as the alternating harmonic series (1/n - 1/(n+1)). Overall, the rate of divergence for a series can vary greatly depending on the specific terms and functions involved.

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