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

AI Thread Summary
The discussion focuses on proving the inequality involving the series u_{n} = ∑_{k=1}^{n} (1/(n+√k)), specifically showing that n/(n+√n) ≤ u_{n} ≤ n/(n+1). Initial attempts at using induction were deemed ineffective due to difficulties in transitioning from n to n+1 in the denominator. A participant suggests bounding the series by finding simpler upper and lower bounds for the terms in the series. The conclusion emphasizes that a more straightforward approach to bounding the series can lead to a successful proof of the inequality.
zodian
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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

Thanks in advance :)
 
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What is \sum_{k=1}^{n}\frac{1}{n+1} and how can you be sure it is greater than un?
 
\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! :)
 
zodian said:
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

Thanks in advance :)

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
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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