A question related to limit and summation

Question

Suppose that we know $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\bigg(\dfrac{n}{n-i}\bigg)^{\gamma} = c$$ where $c$ is a constant. According to this formula, we can say that: for large $n$, $$\frac{1}{n}\sum_{i=1}^{n}\bigg(\frac{n}{n-i}\bigg)^{\gamma}$$ is a constant.

Can we say that: for large $n$, $\sum_{i=1}^{n}(\frac{n}{n-i})^{\gamma}$ is equal to $n\times c$?

Answer 1

Basically you have$\lim\limits_{n \to \infty} \frac{1}n f(n)=c$. If you calculate the limit w.r.t $n$ it is not allowed that $n$ appears on the RHS.

$$\lim\limits_{n \to \infty} f(n)\neq n\cdot c$$

Answer 2

No. Let $t_n = \frac{1}{n}\sum_{i=1}^{n}(\frac{n}{n-i})^{\gamma}$. Then, by your hypothesis, $\lim_{n \to \infty} t_n = c$ for some $c \in \mathbb R$. Most likely, for some large $N$, $$ t_N < t_{N+1} < t_{N+2} < \cdots < c$$