The sequence $(a_n), \ a_1=1, a_{n+1}=2+1/n+3/a_n$ converges The sequence $(a_n), \ a_1:=1, a_{n+1}:=2+1/n+3/a_n$ converges.
Proof.
Let $a_{k-1}\geq 3$ for some positive integer $k$. Then $a_k\leq3+\frac{1}{k-1}=\frac{3k-2}{k-1}$. And $a_{k+1}\geq 2+1/k+\frac{3k-3}{3k-2}$. It is easy to check that $1/k+\frac{3k-3}{3k-2}>1$ for $k>1$. But $a_2=6$, $a_3=3$ and $3\leq a_n$ is proved by induction for positive inthegers greater than $1$.
Now consider a sequence $(x_i)_{i=2}^\infty$, $x_{n+1}:=3+3/n$. Notice that $a_{n+1}=\frac{2n+1}{n}+3/a_n\leq \frac{2n+1}{n}+1$. But $x_{n+1}$ may be rewritten as $\frac{2n+3}{n}+1$. Therefore $x_n\geq a_n$.
And by the sandwich theorem we conclude that $\lim a_n=3$.
P.s. English is not my first language, so it will be excellent to get some hints about grammar, especially articles.
 A: Clearly $a_n>2$, for $n>1$ and hence $3/a_n<3/2$ and thus $a_{n+1}=2+1/n+3/a_n=4$. Hence
$$
2<a_n \le 4, \quad \text{for}\,\,n>2.
$$
Next
$$
|a_{n+2}-a_{n+1}|=
\left|
\frac{1}{(n+1)(n+2)}
+\frac{3(a_{n}-a_{n+1})}{a_na_{n+1}}\right|\le
\frac{1}{(n+1)(n+2)}
+\frac{3|a_{n}-a_{n+1}|}{a_na_{n+1}} \\
\le
\frac{1}{(n+1)(n+2)}
+\frac{3}{4}|a_{n}-a_{n+1}|
$$
So, if $b_n=|a_{n+1}-a_n|$, then
$$
0\le \frac{b_{n+1}}{\left(\frac{3}{4}\right)^{n+1}}-\frac{b_n}{\left(\frac{3}{4}\right)^{n}}\le \frac{1}{(n+1)(n+2)\left(\frac{3}{4}\right)^{n+1}}
$$
and hence
$$
\frac{b_n}{\left(\frac{3}{4}\right)^{n}}\le \frac{3}{4}|a_1-a_2|
+\sum_{k=1}^{n-1}\frac{1}{k(k+1)\left(\frac{3}{4}\right)^{k}}
\,\,\Longrightarrow\,\, b_n\le \left(\frac{3}{4}\right)^{n+1}b_1+
\sum_{k=1}^{n-1}\frac{\left(\frac{3}{4}\right)^{n-k}}{k(k+1)}
$$
and hence
$$
\sum_{n=1}^\infty b_n\le \sum_{n=1}^\infty \left(\frac{3}{4}\right)^{n+1}b_1+\sum_{n=1}^\infty\sum_{k=1}^{n-1}\frac{\left(\frac{3}{4}\right)^{n-k}}{k(k+1)}=3b_1+\frac{16}{3}\sum_{k=1}^\infty
\frac{1}{k(k+1)}<\infty
$$
Hence $\sum_{n=1}^\infty |a_{n+1}-a_n|<\infty$, which implies that $\{a_n\}$ converges. Let $a_n\to x$. Then
$$
x \leftarrow a_{n+1}=2+\frac{1}{n}+\frac{3}{a_n}\to 2+0+\frac{3}{x}.
$$
Thus $x=2+3/x$ or $x^2-2x-3=0$. This means that $x=3$ or $x=-1$. But $a_n\ge 2$, for all $n>1$, and finally $x=3$.
