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$$U_{n+1} =\ \frac{2\left( n^{2} +n+1\right) +nU_{n}}{( n+1)^{2}}, \quad \mbox{ for } n\geq 2,$$ 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!

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  • $\begingroup$ Notice that $U_n\geq 0$, since $U_1=1$ and all the terms in he formula are positive. $\endgroup$ – Dog_69 Nov 6 '18 at 11:16
  • $\begingroup$ How did you conclude that U_1=1 ? n > 0 and we don't have U_0 $\endgroup$ – Alae Cherkaoui Nov 6 '18 at 11:23
  • $\begingroup$ 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 $2$. $\endgroup$ – Dog_69 Nov 6 '18 at 11:28
  • $\begingroup$ The exercice's argument is that n is different than 0 $\endgroup$ – Alae Cherkaoui Nov 6 '18 at 11:33
  • $\begingroup$ Then you should write it explicitly in your question $\endgroup$ – Dog_69 Nov 6 '18 at 11:39
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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 $U_{n+1}>U_n$.

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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 $n\ge1$.

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$$U_{n+1}-U_n=\frac{2(n^2+n+1)-U_n(n^2+n+1)}{(n+1)^2}=\frac{(2-U_n)(n^2+n+1)}{(n+1)^2}$$ So if $U_n<2$ then $U_n$ is increasing since $U_{n+1}-U_n>0$ also if $U_n<2$ we have $$U_{n+1}<\frac{2(n^2+n+1)+2n}{(n+1)^2}=2$$

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