increasing union of finitely generated submodules of M need not be finitely generated Show by an example that an increasing union of finitely generated submodules of M need not be finitely generated.
I was thinking about $R[x_1,x_2,x_3,....]$. Then if we consider the ideal $<x_1>$ , does it form a submodule?
$f(x_1,x_2).g(x_1)\notin<x_1>$, where $f(x_1,x_2)\in R[x_1,x_2,x_3,....]$  and $g(x_1)\in <x_1>$.
So it is not closed under multiplication and hence is not a submodule of $R[x_1,x_2,x_3,....]$
Is this correct?
Can you help me to find an answer for this problem?

You don't quite say what $M$ is.  I'm assuming it's a module over a commutative ring $R$.  IF $M$ is required to be finitely generated, then in order to get any non-(finitely generated) submodule of $M$ we need $R$ to be non-Noetherian, and your example seems as simple as any.
On the other hand, you don't say that you need $M$ to be finitely generated.  If you don't require finite generation of $M$, there are simpler examples: take $R = \mathbb{Z}$, $M = \mathbb{Q}$: 
$\mathbb{Z} \subsetneq \frac{1}{2} \mathbb{Z} \subsetneq \frac{1}{6} \mathbb{Z} \subsetneq \ldots \subsetneq \frac{1}{n!} \mathbb{Z} \subsetneq \ldots \subset \mathbb{Q}$.
