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

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Homework Help Overview

The discussion revolves around proving an inequality involving a series defined as \( u_{n} = \sum_{k=1}^{n}\frac{1}{n+\sqrt{k}} \). The inequality to be proven is \( \frac{n}{n+\sqrt{n}} \leq u_{n} \leq \frac{n}{n+1} \). Participants are exploring methods to approach this proof, particularly considering the use of induction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to apply induction due to the variable \( n \) in the denominator. They also mention having multiple attempts but not sharing them for fear of unproductiveness. Another participant questions the sum \( \sum_{k=1}^{n}\frac{1}{n+1} \) and its relationship to \( u_{n} \). Further, one participant suggests bounding \( t_k \) with a simpler quantity \( u_k \) for easier summation.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding bounding terms, but there is no explicit consensus on the method to be used. The original poster remains uncertain about their approach, while others are questioning assumptions and exploring alternative bounds.

Contextual Notes

There is a mention of the original poster's struggle with the series and the application of induction, indicating a potential gap in understanding the series' behavior as \( n \) changes. The conversation reflects a mix of attempts and questions without a clear resolution.

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
 

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