Compute the limit:
$$\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$$
I tried applying the sandwich rule, constructing an upper and lower bound for the sequence, but I can't get the bounds to have the same limit.
Hints?
Best Answer
Let $S_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}$
The $k^{th}$ term = $t_k = \frac{kn}{n^3+k} \leq \frac{kn}{n^3} = \frac{k}{n^2}$, $\forall{k} \geq 1$
Then, $S_n=\sum\limits_{k=1}^{n}t_k \leq \sum\limits_{k=1}^{n}\frac{k}{n^2} = \frac{1}{n^2}\sum\limits_{k=1}^{n}k=\frac{1}{n^2}.\frac{n(n+1)}{2}=\frac{1}{2}+\frac{1}{2n}=U_n$ (let), $\forall{n} \in \mathcal{N}$
$\lim\limits_{n \to \infty} U_n = \frac{1}{2} \Rightarrow \lim\limits_{n \to \infty} S_n \leq \frac{1}{2}$.
Also, we have, $t_k = \frac{kn}{n^3+k} \geq \frac{kn}{n^3+n} = \frac{k}{n^2+1}$, $\forall{k} \leq n$
Again, $S_n=\sum\limits_{k=1}^{n}t_k \geq \sum\limits_{k=1}^{n}\frac{k}{n^2+1} = \frac{1}{n^2+1}\sum\limits_{k=1}^{n}k=\frac{1}{n^2+1}.\frac{n(n+1)}{2}=\frac{1}{2}.\frac{1+\frac{1}{n}}{1+\frac{1}{n^2}}=V_n$ (let), $\forall{n} \in \mathcal{N}$
$\lim\limits_{n \to \infty} V_n = \frac{1}{2} \Rightarrow \lim\limits_{n \to \infty} S_n \geq \frac{1}{2}$.
Hence, we have sandwich, $U_n \geq S_n \geq V_n$, $\forall{n} \in \mathcal{N}$ and $\lim\limits_{n \to \infty} U_n=\lim\limits_{n \to \infty} V_n=\frac{1}{2}$.
The next figure shows the convergence: