# 3.1 Sums & Products

## Summation

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

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

Exercise 2

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 $$p > -1$$.