Determine convergence or divergence $-1+\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{2}+\frac{1}{4^3}+\frac{1}{5^3}-\frac{1}{3}+...$

Question

Determine convergence or divergence $-1+\dfrac{1}{2^3}+\dfrac{1}{3^3}-\dfrac{1}{2}+\dfrac{1}{4^3}+\dfrac{1}{5^3}-\dfrac{1}{3}+...$

This was a question on my undergrad Real Analysis exam. Would it be enough to state divergence given that the negative terms of the harmonic series are present?

Answer 1

Recall that if two series are convergent, then their series-sum converges, too.

Consider the infinite series:

$$0 - \frac{1}{2^3} - \frac{1}{3^3} + 0 - \frac{1}{4^3} - \frac{1}{5^3} + 0 - \cdots $$

This series converges as the zeroes do not affect the limit of its partial sums, and so we essentially have here $-\sum_{n>1} 1/n^3$, which converges (maybe you have seen this called a $p$-series with $p > 2$).

Suppose your original series converged as well, and we will find a contradiction: For if your series converged, we could add it with the convergent series above to get a convergent series:

$$-1 + 0 + 0 - 1/2 + 0 + 0 - 1/3 + \cdots$$

But this last series diverges: It is the (negation of the) harmonic series with some zeroes interspersed. And so our supposition that your series is convergent was incorrect; it diverges.

Answer 2

Not quite. The negative part is the harmonic series, which diverges. The positive part is convergent, and these two facts together imply the divergence to $-\infty$ of the original series. This isn't hard to show and you should give it a go.

Answer 3

If you check for this wiki page, you will understand why rearrange the series will yields very different result in terms of convergence. And actually, if the series is conditionally converge, you could re-arrange the series to get what ever value on $\mathbb R$.

I also want to point to this question on math.SE for Riemann rearrangement theorem, which has extended discussion on this topic over $\mathbb R^n$.