3.1 Sums & Products

There is a number \(a\) such that \(\sum\limits_{i=1}^\infty i^p\) converges to a finite value iff \(p < a\).

1. What is the value of \(a\)?

   Exercise 1

   &nbsp;Text Response&nbsp; Answer:-1



2. Which of the following would be good approaches as part of a proof that this value of \(a\) is correct?

   Exercise 2

   &nbsp;find a closed form for \(\int_1^\infty x^p\: dx \)&nbsp;

   &nbsp;find a closed form for \(\int_1^\infty i^x\: dx \)&nbsp;

   &nbsp;induction on \(p\)&nbsp;

   &nbsp;induction on \(n\) using the following sum: \(\sum\limits_{i=1}^n i^p \)&nbsp;

   &nbsp;compare the series term-by-term with the Harmonic series&nbsp;

   Approach 1 is correct. For \(p \neq -1 \), the indefinite integral is \(\frac{x^{p+1}}{p+1} \).

   • If \(p < -1\), then \(p+1 < 0\), so \(\lim\limits_{x\to\infty} x^{p+1} = 0\), and the definite integral from 1 to \(\infty\) evaluates to \(\frac{-1}{p+1}\). Hence the sum is bounded from above, and since it is increasing, it has a finite limit, that is, it converges.
   • If \(p > -1\), then \(p+1 > 0\), so \(\lim\limits_{x\to\infty}x^{p+1} = \infty\), and the definite integral diverges.
   • If \(p = -1\), the indefinite integral is \(\log x\), which also approaches \(\infty\) as \(x\) approaches \(\infty\), so the definite integral also diverges.

   Approach 4 is incorrect. This would need the ideas from the good approaches to handle the induction step anyway, at which point the induction would be moot.

   Approach 5 is correct. For \(p = -1\), the sum is the harmonic series, which as we know does not converge. Since the term \(i^p\) is increasing in \(p\) for \(i > 1\), the sum will be larger than the harmonic series, and hence also diverges for \(p > -1\).

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