Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

enter image description here

How can I solve the above problem?

I enter it as:

Limit[Sum[Sqrt[1 + i^2/n^2]/n, {i, n}], n -> Infinity]

but this is not working:

enter image description here

share|improve this question
    
Are you sure the sum converges ? Because I am not .. Anyway, that's easily found in the docs. –  Sektor Dec 6 '13 at 11:09
1  
The limit is infinity but Mathematica does not recognize that and thus returns unevaluated. –  Daniel Lichtblau Dec 6 '13 at 22:42
    
For a practical result try manually converging the upper and lower bounds seen here: Plot[Sum[Sqrt[1 + i^2/n^2]/N[n], {i, 1, n}], {n, 1, 20}] –  Chris Degnen Dec 7 '13 at 12:36
2  
I voted to reopen as a recent edit by the OP changed the interpretation of this question. It's not about typesetting, but about solving the limit. –  Sjoerd C. de Vries Dec 7 '13 at 12:41
    
@MichaelE2 Although true in other cases, it's not possible in cases like this when the terms depend on the upper index. Had mathematica been able to evaluate the Sum[..., {i,n}] then the Limit approach would have had a chance to work. (I'm sure you know this, just thought it was worth clarifying the question in general) –  ssch Dec 7 '13 at 18:48

3 Answers 3

In the limit $n\to\infty$ the sum is the integral (it's just the Riemann sum)

$$ \lim_{n\to\infty}\sum_{i=0}^n\frac{1}{n}\sqrt{1+\frac{i^2}{n^2}}=\int_0^1\sqrt{1+x^2}dx $$ where $dx$ is $1/n$ with $n \to \infty$.

Integrate[Sqrt[1 + x^2], {x, 0, 1}]
% // TrigToExp
%% // N
1/2 (Sqrt[2] + ArcSinh[1])
1/Sqrt[2] + 1/2 Log[1 + Sqrt[2]]
1.14779
share|improve this answer
    
Nice observation! –  ssch Dec 7 '13 at 19:06
    
+1 - looks really good. Haven't figured it all out yet though. –  Chris Degnen Dec 8 '13 at 1:09

This sum does not converge since if you drop the $\frac{i^2}{n^2}$ you get the harmonic series which does not converge and since the lower limit does not converge this sum shouldn't converge either.

share|improve this answer
    
its not clear to me what he asking.. but FWIW you can put Infinity as the iterator limit in the sum. –  george2079 Dec 6 '13 at 14:29
    
It appears to converge within limits 1.1465 < x < 1.148, although it might diverge for n beyond 1000. –  Chris Degnen Dec 7 '13 at 18:04
3  
The sum is over i not n, so one does not get the harmonic series –  ssch Dec 7 '13 at 18:25

It doesn't appear to me that Limit and Sum can be combined, but trying some plotting :-

enter image description here

f[n_] := Sum[Sqrt[1 + i^2/n^2]/N[n], {i, 1, n}]
data = Table[{n, f[n]}, {n, 1, 10000, 1}];
ListLinePlot[data]

enter image description here

share|improve this answer
    
You could sample at only integer values, that get rids of the jiggling –  ssch Dec 7 '13 at 18:29
    
I just realised that. Made edit ;-) –  Chris Degnen Dec 7 '13 at 18:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.