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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
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
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

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]]
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.

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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
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}];

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

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