Telescoping exercise with iterations? 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 $$C(x_*)=-\frac12x_*^4\qquad C\left(x^2-\frac12\right)=-x^4-C(x)$$
 A: $A(x)=- f(x)^4 + f(x,2)^4 - f(x,3)^4 + f(x,4)^4 - ...$
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
$c=\frac{1}{8}$
A: 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.
