Came across this problem recently, posed as apparently an IIT JEE exam question (though I haven't been able to find the exact year). This is a horrible question for the exam (which is usually multiple choice with negative marking) because it rewards bad students and punishes the good students. And as expected, the IIT JEE coaching class solution I found for this was utterly wrong. A sad state of affairs indeed.
Here is the gist of the question (the actual question was a bit different, but this is the crux of the question).
Find $$\lim_{n \to \infty} \dfrac{1 + 2^n + \dots + n^n}{n^n}$$
The wrong solution goes as follow
$$\sum_{j=1}^{n} \left(\dfrac{j}{n}\right)^n = \sum_{k=0}^{n-1} \left(\dfrac{n-k}{n}\right)^n$$
$$ = \sum_{k=0}^{n-1} \left(1 - \dfrac{k}{n}\right)^n = \sum_{k=0}^{n-1} e^{-k} = \dfrac{e}{1-e}$$
The step going to $e^{-k}$ is wrong, without some additional justification. In fact, it can be justified using theorems (like monotone/dominated convergence theorems) which is not in the syllabus of IIT JEE. A good student who does not know those theorems will likely be stuck, while a bad student will happily get the right answer by using pretty faulty reasoning.
Below is an attempt to prove that the limit is $\dfrac{e}{e-1}$ using "elementary" means.
The steps are similar till
$$ \dfrac{1 + 2^n + \dots + n^n}{n^n} = \sum_{k=0}^{n-1} \left(1 - \dfrac{k}{n}\right)^n$$
At this point we try to give upper and lower bounds which ultimately give us the limit we seek,
We use $e^{x} \ge 1 + x$ multiple times.
Setting $x = \frac{-k}{n}$ gives us $e^{-k/n} \ge (1 - k/n)$ and thus $$ \left(1 - \dfrac{k}{n}\right)^n \le e^{-k}$$
Setting $x = \frac{t}{1-t}$ gives us $e^{t/(1-t)} \ge \frac{1}{1-t}$, and so $e^{t} \ge (1-t)^{t-1}$ and thus $(1-t)^{1-t} \ge e^{-t}$
Setting $t = \frac{k}{n}$ and rearranging gives us
$$\left(1 - \dfrac{k}{n}\right)^n \ge \left(1 - \dfrac{k}{n}\right)^k e^{-k}$$
By Bernouli's inequality $\left(1 - \dfrac{k}{n}\right)^k \ge \left(1 - \dfrac{k^2}{n}\right)$ and so we have the bounds
$$e^{-k} \ge \left(1 - \dfrac{k}{n}\right)^n \ge \left(1 - \dfrac{k^2}{n}\right) e^{-k}$$
Taking the sum from $k = 0$ to $n-1$, and noticing that $\sum k^2 e^{-k}$ is $O(1)$ and hence $\frac{1}{n}\sum k^2e^{-k} \to 0$ as $n \to \infty$ gives us the result.
