Can any one help me with this.
We have
$$
U_n=\frac{n^3}{n^4+1}+\frac{n^3}{n^4+2}...+\frac{n^3}{n^4+n}
$$
How to prove that, for all $n$:
$$
\frac{n^4}{n^4+n}\le U_n\le \frac{n^4}{n^4+1}
$$
and what the limit of the sequence ?
I've proved that for $U_0$ but I couldn't prove that for $n+1.$
Thanks too much
On
$\begin{array}\\ U_n &=\frac{n^3}{n^4+1}+\frac{n^3}{n^4+2}...+\frac{n^3}{n^4+n}\\ &=\sum_{i=1}^n \frac{n^3}{n^4+i}\\ &=\dfrac1{n}\sum_{i=1}^n \frac{n^4}{n^4+i}\\ &=\dfrac1{n}\sum_{i=1}^n \frac{1}{1+i/n^4}\\ &<\dfrac1{n}\sum_{i=1}^n 1\\ &=1\\ \text{and}\\ U_n &=\dfrac1{n}\sum_{i=1}^n \frac{1}{1+i/n^4}\\ &>\dfrac1{n}\sum_{i=1}^n \frac{1}{1+n/n^4}\\ &>\dfrac1{n}\sum_{i=1}^n \frac{1}{1+1/n^3}\\ &=\frac{1}{1+1/n^3}\\ &>1-\dfrac1{n^3} \qquad\text{since }\dfrac1{1+x} > 1-x \text{ for } 0 < x < 1\\ \end{array} $
Remark that $\frac{n^3}{n^4+1}$ is the largest term in the expression of $U_n$ so you can upperbound the value of $U_n$ by taking $n\cdot\frac{n^3}{n^4+1}$. Similarly for the lower-bound you can remark that $\frac{n^3}{n^4+n}$ is the smaller term in $U_n$.
Now for the limit... wait one sec ;-)
Using the bound just found you can write $$ \lim_{n\to\infty} \frac{n^4}{n^4+n}\leq \lim_{n\to\infty} U_n\leq\lim_{n\to\infty} \frac{n^4}{n^4+1} $$
Hence $\lim_{n\to\infty} U_n=1$.
But there is not induction here...