Dgdrxrt

I used the cauchy criteria, for $p>q$

$$suplimits_xin [a,b]left | S_p(x) -S_q(x)right |=
suplimits_xleft | sumlimits_k=q+1^p(-1)^nfracx^2+nn^2 right |leq \
suplimits_xsumlimits_k=q+1^pleft | fracx^2+nn^2 right |leq
suplimits_xfracx^2+q+1(q+1)^2(p-q)=\
fracb^2+q+1(q+1)^2(p-q)$$

Is what I am doing right till now ?

Good way! Now each time let $p=lflooralpha qrfloor$ with some $alpha>1$
therefore$$sup_xin [a,b]left | S_p(x) -S_q(x)right |le b^2+q+1over (q+1)^2(alpha-1)q$$and by tending $qtoinfty$ as Cauchy's criterion imposes we obtain$$sup_xin [a,b]left | S_p(x) -S_q(x)right |le alpha-1$$since this is true for every $alpha>1$ therefore $$sup_xin [a,b]left | S_p(x) -S_q(x)right |to 0$$and the proof is complete.

An alternative way

Note that both$$sum_n=1^infty(-1)^nover n=ln 2\sum_n=1^infty(-1)^nover n^2$$are convergent, therefore $$sum_n=1^infty(-1)^nfracx^2+nn^2$$ is uniformly convergent by investigating $sum_n=N^infty(-1)^nfracx^2+nn^2$ for large enough $N$. Another way to say it is that $$|f_n(x)-f(x)|\=
left$$since both $sum_n=N^inftyfrac(-1)^nn$ and $sum_n=N^inftyfrac(-1)^nn^2$ tend to zero and $maxa^2,b^2$ is bounded, then $|f_n(x)-f(x)|$ can be bounded enough and hence the proof is complete.

Since the series is not absolutely convergent, using the bound,

$$left|sum_n=p+1^q (-1)^nfracx^2+nn^2right| leqslant sum_n=p+1^qfracx^2+nn^2,$$

you will not succeed in proving that partial sums of $sum_n geqslant 1(-1)^nfracx^2+nn^2$ satisfy the Cauchy criterion uniformly because the RHS is the difference of partial sums for a divergent series. Note that $fracx^2+nn^2 = mathcalOleft(frac1n right)$.

For the bounding series to satisfy the Cauchy criterion, we must have for any $epsilon > 0$,

$$sup_x in [a,b]sum_n=q+1^p fracx^2+nn^2 < epsilon $$

for all sufficiently large $q$ and $p > q$.

However,

$$sup_x in [a,b]sum_n=q+1^p fracx^2+nn^2 geqslant sup_x in [a,b](p-q) fracx^2 +q p^2 = (p-q) fracmax(a^2,b^2) +q p^2 $$

Taking $p = 2q$, we have

$$sup_x in [a,b]sum_n=q+1^p fracx^2+nn^2 geqslant q fracmax(a^2,b^2) +q 4q^2 = frac1 + max(a^2,b^2)/q4,$$

and since the RHS converges to $1/4$ as $q to infty$, the Cauchy criterion is violated.

A correct approach

Uniform convergence can be established using Dirichlet's test. The hypotheses are met since the partial sums $sum_n=1^N(-1)^n$are uniformly bounded and it can be shown that as $n to infty$, we have

$$fracx^2 + nn^2 downarrow 0,$$

where the convergence is monotonic and uniform for $x in [a,b]$.

The uniform convergence is easily established since,

$$fracmin(a^2,b^2)+nn^2 leqslant fracx^2+nn leqslant fracmax(a^2,b^2) +nn^2$$

See if you can finish by showing that $fracx^2 + nn^2$ is decreasing with respect to $n$.

Summation by parts is another approach, but that in essence repeats the steps in a proof of Dirichlet's test for uniform convergence.