Telescoping exercise with iterations?

Question

For every $x>0$, consider the sequence $(x_n)$ defined by $x_0=x$ and, for every $n\geqslant0$, $$x_{n+1} = \sqrt{x_n + \frac12}$$ Then $x_n\to x_*=\frac{1+\sqrt3}2\ne0$ hence the sequence $$S_n(x)=\sum_{k=1}^n(-1)^kx_k^4$$ diverges. Consider its Cesàro sums, defined by $$C_n(x)=\frac1n\sum_{k=1}^nS_k(x)$$

The question is to prove that $C_n(x)\to C(x)=\frac18-x^2$.

One can probably use telescoping and / or differentiation techniques.

As safety checks, note that the proposed limit $C(x)$ satisfies the relations $$C(x_*)=-\frac12x_*^4\qquad C\left(x^2-\frac12\right)=-x^4-C(x)$$

Answer 1

$A(x)=- f(x)^4 + f(x,2)^4 - f(x,3)^4 + f(x,4)^4 - ...$

The derivative of $A$, with respect to $x$ is

$B(x)=- 4f(x)^3f'(x) + 4f(x,2)^3f'(x,2) - 4f(x,3)^3 f'(x,3)+...$

It can be proved that

$f'(x,n)=f'(x,n-1)\times \frac{1}{2f(x,n)}$

Also, the first term in $B(x)$ can be rewritten as

$- 4f(x)^3f'(x)=-2f^2(x,1)f'(x,0)$

Using the two new relations, we get

$B(x)=-2(f^2(x,1)f'(x,0)-f^2(x,2)f'(x,1)+f^2(x,3)f'(x,2)-...)$

Writing the same thing, in a compact way

$B(x)=-2\sum_{n=0}f^2(x,2n+1)f'(x,2n)+2\sum_{n=1}f^2(x,2n)f'(x,2n-1)$

Then, another relation helps. Substituting $f^2(x,n+1)=f(x,n)+\frac{1}{2}$, in the last equation, gives

$-\sum_{n=0}f'(x,2n-1)-\sum_{n=0}f'(x,2n)+\sum_{n=1}f'(x,2n-2)+\sum_{n=1}f'(x,2n-1)$

Now, consider the first and the last summations together and the summations in the middle together to have

$B(x)=-f'(x,-1)$

The iterative relation gives

$f(x,0)^2=f(x,-1)+\frac{1}{2}$

Therefore

$f(x,-1)=x^2-\frac{1}{2}$

Now, having $f'(x,-1)=2x$

$A'(x)=B(x)=-2x$

Therefore

$A(x)=-x^2+c$

Now, to find the constant $c$, you may notice a trick

$A(0)=-A(-\frac{1}{2})$

which is

$c=-c+\frac{1}{4}$

Finally

$c=\frac{1}{8}$

Answer 2

Update 18.10.2017: In the previous version of my answer I'd assumed the series begins at $x_0$ ; instead of $-x_1$ as it was defined in the question. I adapted now my results and matrices accordingly.

---

Giving examples to questions in the comments.
I used Pari/GP; there is a Cesaro-sum-compatible procedure in it sumalt() With this I got the following table:

f(x)=sqrt(x+1/2)
p=4                      \\ used as exponent for the series
list=vectorv(20)         \\ takes 20 solutions   
\\ --------- put the following commands of the loop in a "bracketed block"
{ for(q=1,20,  x0=x1=q-1;  
       su=sumalt(k=1,
              (-1)^k * ( x1 = f(x1))^p); \\ for iteration (k>0)
       list[q] = [ x0 , su , 1.0/8 - x0^2  ]
    ); }
  \\ ---------
  printp(Mat(list)); 

...

  x0   !  sum by"sumalt" !  1/8 -x0^2         
       !   (cesarosum)   !  equals "sum"
  -----+ ----------------+ ------------------
     0   0.125000000000   0.125000000000
     1  -0.875000000000  -0.875000000000
     2   -3.87500000000   -3.87500000000
     3   -8.87500000000   -8.87500000000
     4   -15.8750000000   -15.8750000000
     5   -24.8750000000   -24.8750000000
     6   -35.8750000000   -35.8750000000
     7   -48.8750000000   -48.8750000000
     8   -63.8750000000   -63.8750000000

The differences are in the digits near software-epsilon

---

[Update] There are some more examples of interesting sums, see the following answer in another MSE-thread

---

There is also an empirical/a heuristic/conjectured result using my earlier discussed matrix-approach, which employs Carleman-matrices and for the summation of the alternating iteration series -when this is possible- the Neumann-series of that Carleman matrix.
My Pari/GP-tools give me the following

pc_f=polcoeffs(f(x),32)~   \\ put the (leading) coefficients of the
                           \\ powerseries-expansion of f(x) into a vector.
                           \\ Because of finite size we can always -at best- 
                           \\ get approximative solutions  
\\ show the first few coefficients:(actually I work with first 32 coeffs.)
[0.707106781187, 0.707106781187, -0.353553390593, 0.353553390593,        
 -0.441941738242, ... ]

F = mkCarlemanmatrix(pc_f)   \\ user defined procedure 

\\The top-left of F is

  1   0.707106781187  0.500000000000   0.353553390593  0.250000000000   0.176776695297
  0   0.707106781187   1.00000000000    1.06066017178   1.00000000000   0.883883476483
  0  -0.353553390593               0   0.530330085890   1.00000000000    1.32582521472
  0   0.353553390593               0  -0.176776695297               0   0.441941738242
  0  -0.441941738242               0   0.132582521472               0  -0.110485434560
  0   0.618718433538               0  -0.132582521472               0  0.0662912607362

With a vector $\small V(x)=[1,x,x^2,x^3,...]$ we can then evaluate the series by doing the dotproduct

  V(x) * F = [1, f(x), f(x)^2, f(x)^3, f(x)^4 , ... ]
            = V(f(x)) 

Here the series which occur at odd indices ($\small f(x),f(x)^3,f(x)^5$) have convergence-radius $\small \rho<=1$
Note, that in the fifth column (=at column with index 4) we get the 4'th power of the function: $\small f(x)^4$ which is of course what shall interest us below for our problem.

---

Now we try to get a meaningful matrix A by the Neumann-Series, which interprets the alternating geometric-series for matrices (as far as this is at all possible). Because your series begins at $x_1$ and not $x_0$ I omit the first term which would be the identity-matrix $F^0$:

$$\small A = - F+ F^2 - F^3 ... = F*(I + F)^{-1} $$

That matrix-inversion must be done with much care; in such cases as here I use LDU-decomposition, (exact(!)) inversion of the components and construction of the inverse from the product of inverses of the L,D,U components applying Eulersummation when convergences in the dotproducts are bad. But even the naive inversion-procedure in Pari/GP gives a seemingly meaningful approximate solution:

  A = -F * (matid(32) + F)^-1 

 \\ the top-left of A
     -1/2   -1.3477094     1.0977094    -0.57076948  0.12500000    0.11653321
        0   -1.0048085  0.0048084741  -0.0024627863           0  0.0013720680
        0    9.3831760    -9.3831760      4.4965232  -1.0000000   -0.94229215
        0  -0.36697128    0.36697128     -1.1877337           0    0.10440858
        0   -5.4154218     5.4154218     -3.2908264           0    0.49966112
        0   -3.9408033     3.9408033     -2.0077436           0    0.11044718
        0    2.4551648    -2.4551648      1.4308645           0    -1.1883271
        0    1.5382960    -1.5382960     0.88397114           0   -0.55144081
        0   -3.4084780     3.4084780     -1.6751387           0    0.95787115
        0   -2.8478962     2.8478962     -1.3608389           0    0.80723231

The dotproducts with a $V(x)$-vector should give approximately: $$ V(x) \cdot_{\mathfrak E} A = [ a_0 , a_1(x), a_2(x) , a_3(x), a_4(x), ... ] $$ The ${\mathfrak E}$ means here that possibly Eulersummation is involved as far as the occuring summations are not convergent, or converge only badly.

But we have two interesting columns here: they have only finitely many entries so that the alternating series for that exponents might be computable by finite polynomials:

  column     represents                           gives value
                                                  by evaluation
  -----------------------------------------------------------------
   a_0    =  - x_1^0 + x_2^0 - ... + ...      =  -1/2
   a_4(x) =  - x_1^4 + x_2^4 - ... + ...      =   1/8 -1*x^2   

The value for $\small a_0$ complies with the evaluation of $\small -1+1-1 \ldots$ by Eulersummation and the values for $\small a_4(x)$ comply with the results gotten by series-summation $\small -x_1^4 + x_2^4 - ... + ...$

---

Such heuristics by the Neumann-matrices can be found in many places in tetration and iteration-series, however I had never time and energy to sit down and do the formal proofs for that concluded properties....

Answer 3

Here is the complete proof. I had wanted to post this before but had to wait for all close voters to rescind their votes after trying to save this question.

First off -- I think the key step in the proof is made a little clearer by using the iterated function notation for the question instead of the other that has been put to use now -- so we will first start off with a rephrase as follows: Let $f(x) = \sqrt{x + \frac{1}{2}}$. Then $x_n = f^n(x)$ and we want to prove that

$$\sum_{k=1}^{\infty} (-1)^k x_k^4 \stackrel{\mathrm{pseudo}}{=} \frac{1}{8} - x^2$$

where the left is divergent but reinterpreted using Cesaro summability (hence the pseudo-equality), i.e. that

$$\lim_{n \rightarrow \infty} C_n(x) = \frac{1}{8} - x^2$$

with $$C_n(x) = \frac{1}{n} \sum_{k=1}^{n} \left(\sum_{l=1}^{k} (-1)^l x_l^4\right)$$.

To do this, first replace $x_l$ by the corresponding iterated functions $f^l(x)$:

$$ \begin{align} C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(\sum_{l=1}^{k} (-1)^l [f^l(x)]^4\right)\\ &= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^1(x)]^4 + [f^2(x)]^4 - \cdots + (-1)^l [f^l(x)]^4\right)\\ \end{align} $$

Now, we note that, by definition of iterated functions that $f^l(x) = f(f^{l-1}(x))$ and this allows us to expand out $[f^l(x)]^4$ as follows:

$$ \begin{align} [f^l(x)]^4 &= [f(f^{l-1}(x))]^4 \\ &= \sqrt{f^{l-1}(x) + \frac{1}{2}}^4 \\ &= \left(f^{l-1}(x) + \frac{1}{2}\right)^2 \\ &= [f^{l-1}(x)]^2 + f^{l-1}(x) + \frac{1}{4} \\ &= [f(f^{l-2}(x))]^2 + f^{l-1}(x) + \frac{1}{4} \\ &= \sqrt{f^{l-2}(x) + \frac{1}{2}}^2 + f^{l-1}(x) + \frac{1}{4} \\ &= f^{l-2}(x) + \frac{1}{2} + f^{l-1}(x) + \frac{1}{4} \\ &= f^{l-2}(x) + f^{l-1}(x) + \frac{3}{4} \end{align} $$

We now plug this back into the previous series to get

$$ \begin{align} C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^1(x)]^4 + [f^2(x)]^4 - \cdots + (-1)^l [f^l(x)]^4\right)\\ &= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^{-1}(x) + f^0(x) + \frac{3}{4}] + [f^0(x) + f^1(x) + \frac{3}{4}] - [f^1(x) + f^2(x) + \frac{3}{4}] + [f^2(x) + f^3(x) + \frac{3}{4}] - \cdots + (-1)^k [f^{k-2}(x) + f^{k-1}(x) + \frac{3}{4}]\right)\\ &= \frac{1}{n} \sum_{k=1}^{n} \left(-f^{-1}(x) - [f^0(x) - f^0(x)] - \frac{3}{4} + [f^1(x) - f^1(x)] + \frac{3}{4} - [f^2(x) - f^2(x)] - \frac{3}{4} + [f^3(x) - f^3(x)] + \frac{3}{4} - \cdots + (-1)^k [f^{k-1}(x) - f^{k-1}(x)] + (-1)^k \frac{3}{4}]\right) \end{align} $$

Thus we now have the series in telescoping form and it telescopes down to

$$\begin{align} C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(-f^{-1}(x) - [k \mod 2 = 1] \frac{3}{4}\right) \end{align} $$

Now of course $f^{-1}(x) = x^2 - \frac{1}{2}$ so

$$ C_n(x) = \frac{1}{n} \sum_{k=1}^{n} \left(-x^2 + \frac{1}{2} - [k \mod 2 = 1] \frac{3}{4}\right) $$

This then becomes three separate Cesaro means

$$ C_n(x) = \left(\frac{1}{n} \sum_{k=1}^{n} (-x^2)\right) + \left(\frac{1}{n} \sum_{k=1}^{n} \frac{1}{2}\right) - \left(\frac{1}{n} \sum_{k=1}^{n} [k \mod 2 = 1] \frac{3}{4}\right) $$

Now we take the limit as $n \rightarrow \infty$. The first two means are means of identical numbers, thus they respectively equal $-x^2$ and $\frac{1}{2}$. The last one is like the mean of Grandi's series, which has limit $\frac{1}{2}$, so this has limit $\frac{3}{8}$. Thus the final Cesaro sum is

$$ \begin{align} C(x) &= \lim_{n \rightarrow \infty} C_n(x)\\ &= -x^2 + \frac{1}{2} - \frac{3}{8}\\ &= \frac{1}{8} - x^2 \end{align} $$.

or

$$\sum_{k=1}^{\infty} (-1)^k x_k^4 \stackrel{\mathrm{pseudo}}{=} \frac{1}{8} - x^2$$

QED.