# Proof of an inequality involving a series (probably by induction)

1. Sep 11, 2012

### zodian

$$u_{n} = \sum_{k=1}^{n}\frac{1}{n+\sqrt{k}}$$
Proof that:
$$\frac{n}{n+\sqrt{n}} \leq u_{n} \leq \frac{n}{n+1}$$

Ok, I've been working on that problem for about two hours now and I still don't have a clue how to proof this inequality.
I guess it should be done by induction, but I have problems with the series, because I don't know how I could possibly pass from n to n+1, since the variable n is on the denominator.
Perhaps there is a pretty easy solution to this problem, but any help would be welcome!
(I'm sorry that I don't post my attempts at a solution, but I have to much of them and I don't believe that there is anything really productive

2. Sep 11, 2012

### Mentallic

What is $$\sum_{k=1}^{n}\frac{1}{n+1}$$ and how can you be sure it is greater than un?

3. Sep 11, 2012

### zodian

$$\sum_{k=1}^{n}\frac{1}{n+1} = \frac{n}{n+1}$$
$$n+\sqrt{k} \geq n+1$$ for every $k\geq 1$
Thus $$\frac{1}{n+\sqrt{k}} \leq \frac{1}{n+1}$$
and $$\sum_{k=1}^{n}\frac{1}{n+\sqrt{k}}\leq \sum_{k=1}^{n}\frac{1}{n+1}$$

And nearly the same works for the other part of the inequality

So I guess I was totally mistaken with tryng to apply induction...
Well thanks anyway! :)

4. Sep 11, 2012

### Ray Vickson

The word you want is "prove", not "proof". Anyway, if $$t_k =\frac{1}{n + \sqrt{k}},$$ can you find a simpler quantity $u_k$ that bounds $t_k$ from above and is easy to sum? That is, can you think of a bound $t_k \leq u_k ,$ where $u_k$ is easier to deal with? Can you do something similar for a lower bound $l_k \leq t_k?$

RGV