1. ## Sequence

Hi everybody,

$U_n=1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+.. .+\frac{1}{n^3}$

I must show that $(\forall n \in \mathbb{N}^*) U_n\le 2-\frac{1}{n}$

I used recurrence but i didn't find anything.

2. Originally Posted by bhitroofen01
Hi everybody,

$U_n=1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+.. .+\frac{1}{n^3}$

I must show that $(\forall n \in \mathbb{N}^*) U_n\le 2-\frac{1}{n}$

I used recurrence but i didn't find anything.
Use induction:

The base case $n=1$ is true.

Assume true for $n=k$, then

$U_{k+1}=U_k+\frac{1}{(k+1)^3}\le 1-\frac{1}{k}+\frac{1}{(k+1)^3}$

and with a bit of work this can be manipulated to show that:

$U_{k+1}\le 2 - \frac{1}{k+1}$

which allows one to claim proof bt induction

CB

3. And how to show that $(U_n)$ is convergent

I know just that $\lim_{n\to +\infty} U_n \le 2 \ \$????

4. The integral test can be used to show that $\sum_{n=1}^\infty \frac{1}{n^r}$ converges for r> 1.

5. Originally Posted by bhitroofen01
And how to show that $(U_n)$ is convergent

I know just that $\lim_{n\to +\infty} U_n \le 2 \ \$????
$U_n$ is an increasing bounded sequence and hence converges, and so the series converges.

CB