# How do you find the positive values of p for which #Sigma n(1+n^2)^p# from #[2,oo)# converges?

and the series is divergent.

Certainly:

fits the purpose, so the series:

will converge if:

also converges.

Let's look at:

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To find the positive values of ( p ) for which the series ( \sum_{n=2}^{\infty} n(1+n^2)^p ) converges, we can use the limit comparison test or the ratio test.

First, let's consider the ratio test. We evaluate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where ( a_n = n(1+n^2)^p ).

[ \lim_{n \to \infty} \left| \frac{(n+1)(1+(n+1)^2)^p}{n(1+n^2)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)(1+n^2+2n+1)^p}{n(1+n^2)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)(n^2+2n+2)^p}{n(1+n^2)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)(n^2(1+2/n+2/n^2))}{n(1+1/n^2)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n(1+1/n^2)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n(1+0)^p} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n} \right| ]

[ = \lim_{n \to \infty} \left| n^2 + 2 + 2/n \right| ]

As ( n \to \infty ), ( n^2 ) dominates the expression, and the limit becomes ( \infty ).

Since the limit is greater than 1, the series diverges for all ( p ) when ( p > 0 ). Therefore, there are no positive values of ( p ) for which the series converges.

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