How to show that the limit of $f(n) \rightarrow 0$ in $n\rightarrow \infty$ I am trying to find a way to show that the following expression goes to $0$ in $n\rightarrow \infty$:
$$
f(n)  = \frac{\frac{(n-d)(n-1)}{n}-\frac{(n-d)(n-1)}{\sum_{i=1}^n \frac{1}{e_i}}}{n(n-1) - \sum_{i=1}^n e_i (n-1)} 
  = \frac{(n-d)\left(\frac{1}{n}-\frac{1}{\sum_{i=1}^n \frac{1}{e_i}}\right)}{n-\sum_{i=1}^n e_i}.
$$
where $d$ is some finite integer constant, $n\in \mathbb Z$, $n>d>0$ and $ 1> e_1\geq e_2\geq ...\geq e_n\geq \frac{1}{n-1}$.
I am not sure how to deal with infinite sums in this problem as I don't think that they converge (?).
Any help would be appreciated.
 A: Consider the Harmonic Mean:
$$HM(e_1,\ldots,e_n) = \dfrac{n}{\displaystyle \sum_{i=1}^n \dfrac{1}{e_i}}$$
and the Arithmetic Mean:
$$AM(e_1,\ldots, e_n) = \dfrac{\displaystyle \sum_{i=1}^n e_i}{n}$$
You have:
$$\begin{align*}f(n) & = \dfrac{(n-d)\left(\dfrac{1}{n}-\dfrac{1}{\displaystyle \sum_{i=1}^n \dfrac{1}{e_i}}\right)}{\displaystyle n-\sum_{i=1}^n e_i} \\ & = \left(\dfrac{n-d}{n^2}\right)\left(\dfrac{1-HM(e_1,\ldots,e_n)}{1-AM(e_1,\ldots,e_n)}\right)\end{align*}$$
For the term on the right, let's look for an upper bound for the numerator and a lower bound for the denominator.
$$1-HM(e_1,\ldots,e_n) \le 1-\min(e_1,\ldots,e_n) = 1-e_n < 1 \\ 1-AM(e_1,\ldots,e_n) \ge 1-\max(e_1,\ldots,e_n) = 1-e_1$$
Thus, $$0 \le f(n) \le \left(\dfrac{n-d}{n^2}\right)\left(\dfrac{1}{1-e_1}\right)$$
And so by the Squeeze theorem, the limit as $n \to \infty$ is $0$. QED.
Based on your comment above, it sounds like some of the $e_i$'s might be equal to one, but you know there exists some $e_i<1$. If that is the case, then the argument changes slightly. There must exist $N\in \mathbb{N}$ such that for all $n\ge N$,
$$AM(e_1,\ldots, e_n) \le AM(e_1,\ldots, e_N) < 1 \\ \Longrightarrow 1-AM(e_1,\ldots, e_n) \ge 1-AM(e_1,\ldots, e_N) > 0$$
And then the proof follows just the same. You just need a constant, so since the $e_i$'s are nonincreasing, you just need a fixed number of them, and the arithmetic mean of those will be constant.
