Xrfgtjtk

I have this series:
$$sum_{n=1}^infty frac{(n)^2(-1)^n} {1+(n^2)} $$

Like every absolutely convergent series are convergent i am working on:
$$sum_{n=1}^infty frac{(n)^2} {1+(n^2)} $$

The ratio test gives me:

$lim_{n to infty} frac{frac{(n+1)^2}{1+(n+1)^2}}{frac{(n)^2}{1+(n)^2}}$ = 1

Wich is inconclusive. Then i take the limit:

$lim_{n to infty} a_n= lim_{n to infty} frac{(n)^2} {1+(n^2)} $= 1

That is different to zero.
So i can conclude that this series diverges .
What do you think? Is this correct?

If a series converges the the lim of the sequence is zero, so this series diverges.

The series is diverge but your proof is not correct because a series which is not absolutely convergent can also be convergent (conditionally convergent).

$$sum_{ngeq 1}frac{(-1)^n n^2}{1+n^2}$$
is neither conditionally or absolutely convergent since the absolute value of the main term does not tend to zero as $nto +infty$. On the other hand it is Cesàro-summable:
$$ lim_{varepsilonto 0^+}sum_{ngeq 1}frac{(-1)^n n^2}{1+n^2}e^{-varepsilon n}=-frac{1}{2}-sum_{ngeq 1}frac{(-1)^n}{1+n^2}=frac{1}{4}left(frac{pi}{sinh pi}-1right). $$
The last identity can be seen as a consequence of the Poisson summation formula, the (inverse) Laplace transform or just Herglotz' trick.

We may rewrite it as $$sum_{n=1}^infty frac{(n)^2(-1)^n} {1+(n^2)}{=-{1over 2}+sum_{n=1}^infty frac{(2n)^2} {1+(2n)^2}-frac{(2n+1)^2} {1+(2n+1)^2}\=-{1over 2}+sum_{n=1}^infty frac{1} {1+(2n+1)^2}-frac{1} {1+(2n)^2}\=-{1over 2}+sum_{n=1}^infty frac{-4n-1} {(1+(2n+1)^2)(1+(2n)^2)}}$$which converges for sure since $$frac{-4n-1} {(1+(2n+1)^2)(1+(2n)^2)}sim -{1over 4n^3}$$for large enough $n$.

It's definitely divergent, that one, because the terms become closer to 1 in magnitude with increasing $n$; the sum becomes more & more like just alternating beween adding & subtracting 1 as in increases.

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With $ds{N in mathbb{N}_{geq 1}}$:

begin{align}
sum_{n = 1}^{N}{n^{2}pars{-1}^{n} over 1 + n^{2}} & =
sum_{n = 0}^{leftlfloor N/2rightrfloor}
{4n^{2} over 1 + 4n^{2}} -
sum_{n = 0}^{leftlfloorpars{N - 1}/2rightrfloor}
{pars{2n + 1}^{2} over 1 + pars{2n + 1}^{2}}
\[5mm] & =
sum_{n = leftlfloorpars{N - 1}/2rightrfloor + 1}^{leftlfloor N/2rightrfloor}
{4n^{2} over 1 + 4n^{2}}
\[5mm] &
+ underbrace{sum_{n = 0}^{leftlfloorpars{N - 1}/2rightrfloor}
bracks{{4n^{2} over 1 + 4n^{2}} - {pars{2n + 1}^{2} over 1 + pars{2n + 1}^{2}}}}
_{ds{stackrel{mrm{as} N to infty}{LARGEto}
-,{1 + picscpars{pi} over 2}}}
end{align}

Note that

begin{align}
&bbox[10px,#ffd]{sum_{n = leftlfloorpars{N - 1}/2rightrfloor + 1}^{leftlfloor N/2rightrfloor}
{4n^{2} over 1 + 4n^{2}}}
\[5mm] = &
left{begin{array}{lcl}
ds{{4pars{m - 1}^{2} over
1 + 4pars{m - 1}^{2}} +
{4m^{2} over 1 + 4m^{2}}} & mbox{if} &
ds{N} even, ds{N equiv 2m}
\[2mm]
ds{4m^{2} over 1 + 4m^{2}} & mbox{if} &
ds{N} phantom{n}odd, ds{N equiv 2m + 1}
end{array}right.
\[5mm] implies &
lim_{N to infty}bbox[10px,#ffd]{sum_{n = leftlfloorpars{N - 1}/2rightrfloor + 1}^{leftlfloor N/2rightrfloor}
{4n^{2} over 1 + 4n^{2}}}
mbox{doesn't} exist.
end{align}

Further remark on Mostafa's answer.
$$
lim_{Ntoinfty}sum_{n=1}^{2N} frac{(n)^2(-1)^n} {1+(n^2)}=
frac{-pi+sinh pi}{2sinh pi} approx 0.36,
\
lim_{Ntoinfty}sum_{n=1}^{2N+1} frac{(n)^2(-1)^n} {1+(n^2)}=
frac{-pi-sinh pi}{2sinh pi} approx -0.64.
$$
These do not agree, so $sum_{n=1}^infty$ does not exist.