Confusion in applying Cauchy's Theorem of limits

Question

Cauchy's theorem of limits states that if $\ \lim_{n \to \infty} a_n=L ,$ then $$ \lim_{n \to \infty} \frac{a_1+a_2+\cdots+a_n}{n}=L $$ If I apply this in the series $$S = \lim_{n\to\infty} \dfrac{1}{n}[e^{\frac{1}{n}} + e^{\frac{2}{n}} + e^{\frac{3}{n}} + e^{\frac{4}{n}} + e^{\frac{5}{n}} + e^{\frac{6}{n}} + ... + e^{\frac{n}{n}}]$$ Here $a_n=e^{n/n},$ Hence, $\lim_{n \to \infty} a_n=e \ $, so $S=e$. I know this is incorrect because the integration gives the answer $e-1$. It seems to me that the reason for this is related to the terms of the sequence being dependent on $n$. I can't be sure of this because the same theorem gives the correct answer for the limit of this series(below) which is given in my book as a solved example of the above theorem $$ \ \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}} \right)=1 $$

Answer 1

The main issue is that your series UP TO $n$ are $a_i=e^{\frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.

Answer 2

You can't apply the theorem in the first case since $$a_n=e^{n\over n}=e$$which doesn't make sense. Another reason is that $a_k=e^{k\over n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$\lim_{n\to \infty}{1\over n}\sum_{i=1}^{n} e^{i\over n}=\int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {n\over \sqrt{n^2+n}}\le\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\le{n\over \sqrt{n^2+1}}$$and use squeeze theorem to attain the result.