I am trying to study the convergence of the sequence defined by:

u0=a>0,un+1=√n∑k=0uku_0 = a>0, u_{n+1} = \sqrt{\sum_{k = 0} ^{n} u_k}

I have shown that un+1−un=unun+1+unu_{n+1} – {u_n} = \frac{u_n}{u_{n+1}+{u_n}} \tag{$*$} since

un+1−un=√n∑k=0uk−√n−1∑k=0uku_{n+1} – {u_n} = \sqrt{\sum_{k=0}^{n}u_k} – \sqrt{\sum_{k=0}^{n-1}u_k}

then rationalize to get to (∗)(*).

And since it can be easily shown that (induction) that un>0u_n>0 for all nn we have that un+1−un>0⇒un+1>unu_{n+1}-u_n>0 \Rightarrow u_{n+1}>u_n

S0 (un)(u_n) is an increasing sequence. Suppose (un)(u_n) was bounded above, then un→uu_n \rightarrow u for some u∈Ru\in\mathbb{R}. Passing to the limit in (∗)(*) we have the u=0u = 0, but this is impossible since u0>0u_0>0 and (un)(u_n) increasing so then (un)(u_n) does not converge.

Is this argument correct?

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Cesaro tells if it converges, it’s to 00.

– Abdallah Hammam

2 days ago

@AbdallahHammam I have not learnt that yet unfortunately, is my argument above correct?

– Dman

2 days ago

@Dman: How did you show un+1−un=unun+1+un?u_{n+1}-u_n=\frac{u_n}{u_{n+1}+u_n}? Furthermore, I’ll agree that un>0u_n>0 for all nn, how are you deducing that un+1−un>0u_{n+1}-u_n>0?

– Clayton

2 days ago

@Clayton, write out the term for un+1u_{n+1} explicitly on the LHS and then rationalize. Since all the terms on the RHS are positive, the LHS must be positive as well.

– Dman

2 days ago

@Clayton see the updated answer

– Dman

2 days ago

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3 Answers

3

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Another way of seeing this is to note that (un)(u_n) also verifies the recurrence formula

un+1=√un+u2nu_{n+1}=\sqrt{u_n+u_n^2}

and study the function f:x↦√x+x2f:x\mapsto \sqrt{x+x^2}. This function stabilizes the interval [0,+∞[[0,+\infty[, is strictly increasing and it is easy to prove that x>0x>0 implies f(x)>xf(x)>x.

So the sequence (un)(u_n) is strictly increasing, and cannot be bounded because, as you said, the only fixed point by ff is 00.

It would be funny then to find an equivalent of unu_n 🙂

This is what I found : from un+1=√un+u2nu_{n+1}=\sqrt{u_n+u_n^2}, you can first derive, as you pointed,

un+1−un=unun+un+1<12u_{n+1}-u_n=\frac{u_n}{u_n+u_{n+1}}< \frac{1}{2}
because (un)(u_n) is strictly increasing. This leads to
un=u1+n−1∑k=1uk+1−uk≤1+n−12=n+12u_n=u_1+\sum_{k=1}^{n-1} u_{k+1}-u_k\le 1+\frac{n-1}{2}=\frac{n+1}{2}
But you can also derive
un+1=un√1+1unu_{n+1}=u_n\sqrt{1+\frac{1}{u_n}}
and because we proved limun=+∞\lim u_n=+\infty, we can use the development :
u_{n+1}=u_n(1+\frac{1}{2u_n}+o(\frac{1}{u_n})) = u_n+\frac12 +o(1)u_{n+1}=u_n(1+\frac{1}{2u_n}+o(\frac{1}{u_n})) = u_n+\frac12 +o(1)
so
u_{n+1}-u_n=\frac{1}{2}+o(1)\sim \frac12u_{n+1}-u_n=\frac{1}{2}+o(1)\sim \frac12
and by Cesaro theorem (or by summations of equivalents) :
u_n = u_1+\sum_{k=1}^{n-1} u_{k+1}-u_k \sim \frac n2u_n = u_1+\sum_{k=1}^{n-1} u_{k+1}-u_k \sim \frac n2
So now you have the limit, an upper bound and an equivalent 🙂
Thanks for this alternative!
– Dman
2 days ago
Since every u_nu_n is positive, we have u_n \ge \sqrt {u_0}u_n \ge \sqrt {u_0} for n \ge 1n \ge 1, and then u_n \ge \sqrt {(n-1)\sqrt {u_0}}u_n \ge \sqrt {(n-1)\sqrt {u_0}} for n \ge 2n \ge 2, which shows that (u_n)(u_n) diverges.
let
v_n=\frac{\sum_{k=0}^n u_k}{n+1}v_n=\frac{\sum_{k=0}^n u_k}{n+1}.
we have
\frac{u_{n+1}^2}{n+1}=v_n\frac{u_{n+1}^2}{n+1}=v_n.
if lim_{n\to +\infty}u_n=Llim_{n\to +\infty}u_n=L then
0=\lim_{n\to+\infty} v_n=L0=\lim_{n\to+\infty} v_n=L
using Cesaro average.
Now, if (u_n)(u_n) is increasing and
u_0=a>0u_0=a>0, the limit can’t be 00.

thus (u_n)(u_n) diverges.

OR WITHOUT CESARO

we have

u_{n+1}-u_n=\frac{u_n}{u_n+u_{n+1}}u_{n+1}-u_n=\frac{u_n}{u_n+u_{n+1}}

and when n\to +\inftyn\to +\infty

L-L=\frac{L}{2L}L-L=\frac{L}{2L}

which proves that

(u_n)(u_n) diverges.