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What I was trying to do, was to prove if $\lim _{n \to \infty} a_n = L$ then $\lim _{n \to \infty} \frac{ \sum_{k=1}^{n}a_k}{n} = L$ is true or not. After some trials and errors, I found the series $a_k=k^{\frac{1}{k}}$ where $$\lim _{n \to \infty} n^{\frac{1}{n}}=1$$ and according to wolframalpha, $$\lim _{n \to \infty} \frac{ \sum_{k=1}^{n}k^{\frac{1}{k}}}{n}=0$$but I doubt this, as $x^{\frac{1}{x}}>1$ for $x>1$ thus $\lim _{n \to \infty} \frac{ \sum_{k=1}^{n}k^{\frac{1}{k}}}{n}\geq1$

I would appreciate proof of the proposition, or counterexample, or an extra reasons on why $\lim _{n \to \infty} \frac{ \sum_{k=1}^{n}k^{\frac{1}{k}}}{n}=0$

Gary
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Saeol Ahn
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    The general assertion follows from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem WolframAlpha is incorrect. – Gary Apr 07 '24 at 00:33
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    What data from Wolfram Alpha makes you believe it? – KCd Apr 07 '24 at 00:34
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    Try typing "integrate sin(pi x)/(x*(1-x)) from 0 to 1" into WA. Sometimes it says $0$, which is absurd since the integrand is positive on $(0,1)$. Also sometimes it says $3.70387$, which is accurate. Whichever answer you get, try asking WA again and you'll get the other answer eventually (it always happens to me). – KCd Apr 07 '24 at 00:38

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Don't say you doubt the answer can be zero in your example, but rather that you know the answer is not zero in your example, as you showed us a simple reason the limit can't be zero in that example: the $n$th term is always at least $1$.

To prove your goal, a nice way to streamline things is to reduce to the case $L = 0$. When $a_n \to L$, set $b_n= a_n - L$, so $b_n \to 0$. Then $$ \frac{\sum_{k=1}^n a_k}{n} - L = \frac{\sum_{k=1}^n (a_k-L)}{n} = \frac{\sum_{k=1}^n b_k}{n}, $$ so proving $(\sum_{k=1}^n a_k)/n \to L$ is equivalent to proving $(\sum_{k=1}^n b_k)/n \to 0$. Thus the general result you want to show is equivalent to the special case $L = 0$, so you may assume $L = 0$.

KCd
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