# proof of upper bound for a sequence convergence: $U_{n+1} =\ \frac{2\left( n^{2} +n+1\right) +nU_{n}}{( n+1)^{2}}$

$$U_{n+1} =\ \frac{2\left( n^{2} +n+1\right) +nU_{n}}{( n+1)^{2}}, \quad \mbox{ for } n\geq 2,$$ and $$n \in \Bbb N\setminus\{0\}$$, $$U_1<2$$.

I have to prove that this sequence converges by finding an upper bound and proving that $$U_n$$ is increasing.
I am unable to prove that $$U_n$$ is increasing.

My attempt:

I tried with $$U_{n+1} - U_n$$, but I can't conclude the sign of this difference since $$U_n$$ can be both positive and negative.

Please, explain my mistake and provide the best approach to this question!

• Notice that $U_n\geq 0$, since $U_1=1$ and all the terms in he formula are positive. – Dog_69 Nov 6 '18 at 11:16
• How did you conclude that U_1=1 ? n > 0 and we don't have U_0 – Alae Cherkaoui Nov 6 '18 at 11:23
• We needn't $U_0$: $$U_1=U_{0+1} = \frac{2(0^2+0+1)+0U_0}{(0+1)^2} = 2.$$ In fact you can see I was wrong, because I ignored the factor $2$. – Dog_69 Nov 6 '18 at 11:28
• The exercice's argument is that n is different than 0 – Alae Cherkaoui Nov 6 '18 at 11:33
• Then you should write it explicitly in your question – Dog_69 Nov 6 '18 at 11:39

It is easy to see, by induction, that $$U_n <2$$ for all $$n$$.
Then we have $$U_n<2= \frac{2(n^2+n+1)}{n^2+n+1}$$, hence $$2(n^2+n+1)> U_n(n^2+n+1)$$, thus
$$2(n^2+n+1)+nU_n > (n+1)^2U_n$$. This gives $$U_{n+1}>U_n$$.
Consider $$u_n=v_n+2$$. \begin{align} v_{n+1} &=\color{#C00}{v_{n+1}+2}-2\\[3pt] &=\frac{2(n^2+n+1)+n(\color{#C00}{v_n+2})}{(n+1)^2}-2\\ &=\frac{nv_n}{(n+1)^2} \end{align} Then $$v_1\lt0$$ and $$v_n\lt\frac{v_n}4\lt v_{n+1}\lt0$$ for $$n\ge1$$.
$$U_{n+1}-U_n=\frac{2(n^2+n+1)-U_n(n^2+n+1)}{(n+1)^2}=\frac{(2-U_n)(n^2+n+1)}{(n+1)^2}$$ So if $$U_n<2$$ then $$U_n$$ is increasing since $$U_{n+1}-U_n>0$$ also if $$U_n<2$$ we have $$U_{n+1}<\frac{2(n^2+n+1)+2n}{(n+1)^2}=2$$