The limit $\lim_{n\to \infty}\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}$ $$\lim_{n\to \infty}\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}$$ is equal to
(a) $0$
(b) $1$
(c) $e$
(d) none of these
How should I deal with such limits? What I know $e >1$, so $$e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}>1+1+\ldots+1$$ ($n$ times), so $$\frac{e+e^{\frac{1}{2}}+e^{\frac{1}{3}}+\ldots+e^{\frac{1}{n}}}{n}>\frac{n}{n}=1.$$
So this limit should not be zero. How to deal further.
Thanks in advance!
 A: You can recognise that this is a Cesaro mean. There are related ideas and theorems, see the Stolz-Cesaro Theorem.

If $(a_n)$ is a convergent real/complex sequence, then $\frac{1}{n}\sum_{j=1}^na_n$ tends to $\lim_{n\to\infty}a_n$.

Many proofs can be found on this site.
As $e^{1/n}\to1$, the limit of these means shall also tend to $1$.
A quick proof sketch: for large $k>K$, $e^{1/k}$ is close to $1$ via a uniform bound. As the denominator $n$ grows, the sum of all terms $<K$ will vanish ($K$ is a finite constant, but $n\to\infty$) and you're left with the tail sum, where $|e^{1/k}-1|<\varepsilon$ for all $k>K$. Taking a sum, you can make some fairly simple bounds and deduce the limit exists and is equal to $1$.
A: Let's take @Gary:'s hint but in a pedestrian way:
First of all, we have Bernoulli's inequality
$$(1 + x)^n > 1 + n x$$
for $x > 0$, and $n > 1$. Therefore we conclude
$$1+ x > ( 1 + n x)^{\frac{1}{n}}$$
or
$$1 + \frac{y}{n}  > (1+y)^{\frac{1}{n}}$$
for $y> 0$ and $n> 1$ natural. In particular
$$1 + \frac{e-1}{n} > e^{\frac{1}{n}} >1$$
for $n > 1$ (the RHS since $e>1$). Now sum all these inequalities and get
$$1 + (e-1) \frac{1+\frac{1}{2} + \cdots + \frac{1}{n}}{n} > \frac{e + e^{\frac{1}{n}} + \cdots +e^{\frac{1}{n}}}{n} > 1$$
Now we only have to show that
$$\frac{1+\frac{1}{2} + \cdots + \frac{1}{n}}{n}$$ has limit $0$.  How to proceed?  Let's show for instance that for $n$ large enough the terms are $\le \frac{1}{7}$.  I claim that this happens for $n\ge 14^2$. Indeed, the expression can be written as
$$\frac{1+ \frac{1}{2} +\cdots + \frac{1}{14} }{n} + \frac{\frac{1}{15} + \cdots + \frac{1}{n}}{n}$$
Now, the first term has in the numerator $14$ terms $\le 1$, so the first fraction $\le \frac{14}{14^2} = \frac{1}{14}$.  The second fractions has in the numerator $n-14$ terms $< \frac{1}{14}$, and the denominator $n$. We conclude that the second fraction $< \frac{1}{14}$. Now add them up.
Now, instead of $\frac{1}{7}$, we could reason with $\frac{1}{m}$.
