MHB Understanding the Inequality for Solving Limits with Exponential Terms

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The limit discussed involves evaluating the expression as n approaches infinity, leading to the conclusion that it equals ln(a). The inequality used, which states that a^n is less than or equal to the sum of exponential terms and that this sum is also less than or equal to n times a^n, is central to the solution. While the first two parts of the inequality are considered obvious, the third part, which asserts that the sum of the terms does not exceed n times a^n, requires further clarification. It is noted that there are n terms in the sum, each bounded by a^n, which helps in understanding the inequality's validity. The discussion emphasizes the importance of recognizing such inequalities when solving limits involving exponential terms.
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Hello!

$$\lim_{n\rightarrow \infty }\frac{1}{n}ln(a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}} ), \ a>1$$
I solved the limit by using the following inequality:
$$a^{n}\leq a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}\leq n\cdot a^{n}$$
After I applied a $ln$ and $1/n$ I got $lna$
My question is about that inequality.Where does this come from ?How can I prove it ?Should I notice something about the exercise to know I've to use this inequality?
Thanks!
 
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Isn't this inequality obvious?
 
$$a^{n}\leq a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}$$ this I can see is true, it's obvious
$$a^{n}\leq n\cdot a^{n}$$ like the first one, I can see it's true
$$a^{\frac{n}{1}}+a^{\frac{n}{2}}+...+a^{\frac{n}{n}}\leq n\cdot a^{n}$$ this one,I can't see "how it's true", it's not so clear for me why this is true.
 
There are $n$ terms in the sum, and each does not exceed $a^n$.
 
I understood!
Thank you for your help! :)
 

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