How would I go about calculating $$\lim_{n\to\infty}\frac{\left(1 + \frac11\right)^1 + \left(1 + \frac12\right)^2 + \left(1 + \frac13\right)^3 + \cdots + \left(1 + \frac1n\right)^n}n$$ and $$\lim_{n\to\infty} \frac n{\left(1 + \frac11\right)^1 + \left(1 + \frac12\right)^2 + \left(1 + \frac13\right)^3 + \cdots + \left(1 + \frac1n\right)^n}$$
These limits have to do with $e$ but I don't know how to begin with them. Thanks for any help.
Since $\lim_{n \to \infty} \Bigl(1+\frac{1}{n}\Bigr)^n = e$, the upper limit is just $e$ also, and the lower limit, being its reciprocal, is $1/e$.
Note that for any $\varepsilon > 0$, you can find an $N$ such that for all $k > N$, $\Bigl(1+\frac{1}{k}\Bigr)^k$ is within $\varepsilon/2$ of $e$. If the first $N$ terms have an average that differs from $e$ by $m\varepsilon$, for some $m$, then you can simply tack on another $2mN$ terms to get the average within $\varepsilon$ of $e$. This establishes that the limit of the average is $e$, also.
The foregoing is by no means a rigorous demonstration, but perhaps you will find it convincing enough.
