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