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

What is the value of \(a\)?

Which of the following would be good approaches as part of a proof that this value of \(a\) is correct?
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 \(p > 1\).