Riemann $\zeta(3)$ convergence with Cauchy

Question

I'm an undergraduate freshman math student, and we were asked to prove that the sequence $a{_n} =\sum_{k=1}^{n} \frac{1}{k^3}$ converges (obviously, we weren't asked to calculate its limit.) Our teacher hinted to prove that it's a Cauchy sequence. We don't know much, only Cauchy, several sentences about sequences and limits and monotonic sequences and such (basic first few months of undergraduate freshman). I'm stuck. any hints / ideas?

Here's my attempt:

Let $\varepsilon > 0$. We need to find N, such that for all $m > n > N$, $a_{m}-a_{n} < \varepsilon$. $a_{m}-a_{n} = \sum_{k = n+1}^{m} \frac{1}{k^3}$.

$\sum_{k = n+1}^{m} \frac{1}{k^3} < \frac{m-n}{(n+1)^3}$. But this leads nowhere.

Note: We don't have to prove it by Cauchy, any solution (from the little we have learnt) will do.

Answer 1

For $k\geq 2$ we have $k^2\geq k+1$

and

$$\frac{1}{k^3}\leq \frac{1}{k(k+1)}$$

but

$$\sum_{k=2}^n\frac{1}{k(k+1)}=\sum_{k=2}^n (\frac{1}{k}-\frac{1}{k+1})$$

$$=\frac{1}{2}-\frac{1}{n+1}\leq \frac{1}{2}$$

thus the sequence of partial sums

$S_n=\sum_{k=2}^n\frac{1}{k^3}$ is increasing and bounded, and therefore convergent.

Answer 2

Since $ b_n= \dfrac{1}{n^a} > 0 \space \forall\ n \in\mathbb N$ and $b_n>b_{n+1}$ $$\sum b_n \sim \sum2^nb_{2^n} \rightarrow \sum \dfrac{1}{n^a}\sim \sum(\dfrac{2} {2^a})^n$$ which converges for $a>1$

Answer 3

Hint

$$\frac{1}{n^3} < \frac{1}{n(n-1)(n-2)} \\ \frac{1}{n(n-1)(n-2)}=\frac{1}{2} \left( \frac{1}{(n-2)(n-1)}- \frac{1}{(n-1)n} \right)$$

By combining these it is easy to show that your sequence of partial sums is Cauchy

Answer 4

Using partial fractions and telescoping from your setup, after using the inequality $$k^3=k(k^2)>k(k^2-1)=k(k-1)(k+1):$$

$ \begin{align*}\sum_{k=n+1}^m \frac{1}{k^3}&<\sum_{k=n+1}^m \frac{1}{k(k-1)(k+1)}\\ &=\sum_{k=n+1}^m \frac{1/2}{k-1}-\frac1k+\frac{1/2}{k+1}\\ &=\frac{1/2}{n}-\frac{1/2}{n+1}-\frac{1/2}{m}+\frac{1/2}{m+1}, \end{align*} $

which you can bound above with $\dfrac1N$. (Inspiration came from the usual telescoping way to show that partial sums of $\sum\frac1{n^2}$ are bounded, which could also be adapted to show they are Cauchy.)

Answer 5

The integral test shows us that it converges and that

$$\sum_{n=1}^\infty\frac1{n^3}<1+\int_1^\infty\frac1{x^3}\ dx=\frac32$$

And we also know that it is greater than the first term, since the partial sums are monotonically increasing, so

$$1<\sum_{n=1}^\infty\frac1{n^3}<\frac32$$