# How do you find the positive values of p for which #Sigma lnn/n^p# from #[2,oo)# converges?

The series:

is convergent for

A necessary condition for a series to converge is that its general term is infinitesimal, that is:

Clearly if such limit exists it must coincide with (1).

also converges.

Solving the indefinite integral by parts:

So that:

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To find the positive values of ( p ) for which the series (\sum_{n=2}^{\infty} \frac{\ln n}{n^p} ) converges, we use the integral test. We integrate the function ( \frac{\ln x}{x^p} ) from 2 to infinity and check for convergence.

Let ( u = \ln x ). Then, ( du = \frac{1}{x} dx ). Substituting these into the integral:

[ \int_{2}^{\infty} \frac{\ln x}{x^p} dx = \int_{\ln 2}^{\infty} u \cdot u^{1-p} du = \int_{\ln 2}^{\infty} u^{2-p} du ]

This integral converges if ( 2 - p > 1 ) (since it's a power function). So, ( p < 1 ).

Therefore, the positive values of ( p ) for which the series converges are ( 0 < p < 1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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