For what values of p does the series converge?

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SUMMARY

The series Σn=2 1/(n^p)(ln n) converges for values of p greater than 1. The comparison test is utilized to establish this, comparing the series to the p-series 1/n^p, which is known to converge under the same condition. The presence of the natural logarithm factor, ln n, is negligible in terms of its effect on convergence, effectively behaving as if it adds zero to the exponent p. Therefore, the conclusion is that the series converges for p > 1.

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Homework Statement


For what values of p does the series converge?


Homework Equations


Σn=2 1/(n^p)(ln n)


The Attempt at a Solution


So far, all that is available to me is the integral test, the comparison test and the limit comparison test.

So using the comparison test. 1/(n^p)(ln n) < 1/n^p

1/n^p is a p series and only converges when p>1. So it converges.

But this doesn't feel right :(
 
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limonysal said:
So using the comparison test. 1/(n^p)(ln n) < 1/n^p

1/n^p is a p series and only converges when p>1. So it converges.

But this doesn't feel right :(

This is fine: the comparison of terms is valid for n > 1 (so the summation in the series would have to start at n = 2).

In having looked at a fair number of these p-series related problems, I've come to think of the contribution of a (ln n) factor as equivalent to adding zero to the p exponent when conducting the "p-test"; the natural logarithm in such series seems to make a negligible contribution to the convergence of the series.
 
ah i meant to add in so it converges for p>1

its just that all the p series problems usually end up with p>1 XD
 
limonysal said:
ah i meant to add in so it converges for p>1

its just that all the p series problems usually end up with p>1 XD

That's largely because of the problems authors tend to choose for courses making a first pass through material on infinite series. Not all p-series one may encounter turn out that way... ;-)
 

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