real analysis - A sequence defined by setting $a_{n+1}$=$frac{a_{n}^{p-1}+frac{a}{a_{n}^{p-1}}}{p}$ Find $Lim_{nrightarrowinfty}$ $a_{n}$

Question For p$\in\mathbb{N}$, $a$$>$0 and $a_{1}>0$ define

the sequence $\left\{ a_{n}\right\}$by setting

$a_{n+1}$=$\frac{a_{n}^{p-1}+\frac{a}{a_{n}^{p-1}}}{p}$ Find $Lim_{n\rightarrow\infty}$$a_{n}$

My approach Actually i can prove that the sequence is monotonically
decreasing and bounded below by $\sqrt[p]{a}$

Book mentions the answer $\sqrt[p]{a}$

but i can not say that every monotonically decreasing sequence converges
to its lower bound .There can exist $l$ such that $l\geq$ $\sqrt[p]{a}$

and $Lim_{n\rightarrow\infty}$$a_{n}$= $l$. We can not be sure
about it

it would be wrong if i say $l$=${\sqrt[p]{a}}$

My proof of lower bound using AM $\geq$GM inequality

$\frac{A+B}{p}\geq$$\sqrt[p]{A.B}$

$a_{n+1}$$\geq$$\sqrt[p]{a^{p-1}\frac{a}{a^{p-1}}}$$\Longrightarrow$$a_{n+1}$$\geq$$\sqrt[p]{a}$

Edit Actually Real question in the book was $\Longrightarrow$

And its answer $\Longrightarrow$

I thought there is some misprinting in the book so i corrected it and asked the question.

Answer

Thought it'd be a good idea to write the comment as an answer since I wasn't descriptive enough:

$1)$ Use Monotone Convergence Theorem: It says that bounded increasing (decreasing) sequence converges to its supremum (infimum).

$2)$ You've shown that the sequence is monotone and bounded, hence the infimum exists, hence the limit exists.

$3)$ Let that limit be $l$. Since the limit of any subsequence of a convergent sequence is that same as the limit of the sequence itself, we get the following:

$$lim_{n\rightarrow\infty}a_n=lim_{n\rightarrow\infty}a_{n+1}$$

Hence using the definition of a sequence and taking limits of both sides we get the equation:

Note: This equation is valid only for the formulation of the problem before the edit

$$l=\frac{l^{p-1}+\frac{a}{l^{p-1}}}{p}$$

I suppose you can solve this, if not, I can be of assistance

The problem is your AM-GM: you have the sum of only $2$ things:

Hence

$$a_{n+1}p=a_n^{p-1}+\frac{a}{a_n^{p-1}} \geq 2\sqrt{a}$$

hence $a_{n+1}\geq \frac{2\sqrt{a}}{p}$