$\lim_{n\to\infty} n^{x}(a_{1}a_{2}\ldots a_{n})^{\frac{1}{n}}=ae^x$

Question

Let $ (a_{n})$ be positive sequence, $a,x \in R \quad $ and $ \lim_{n\to\infty} n^{x}a_{n}=a$.

Prove that $\lim_{n\to\infty} n^{x}(a_{1}a_{2}\ldots a_{n})^{\frac{1}{n}}=ae^x$

I know that $\lim_{n\to\infty} (a_{1}a_{2}\ldots a_{n})^{\frac{1}{n}}=\lim_{n\to\infty} a_{n}$ but don't have idea how to use it

Answer 1

HINT

By ratio root criteria we have

$$\frac{(n+1)^{x(n+1)}a_{1}a_{2}\ldots a_{n+1}}{n^{xn}a_{1}a_{2}\ldots a_{n}}=(n+1)^xa_{n+1}\left(1+\frac1n\right)^{nx}$$