Investigate convergence for series (conditionally or/and absolutely) i have to invistigate this series (conditionally or/and absolutely).
$$\sum_{n=1}^\infty{(-1)^n\frac 1{\sqrt n}}$$
so i tried to solve it and this what i did:
let $U_n=\frac {(-1)^n}{\sqrt n}$
$$|\sum{U_n}|=|\sum\frac {(-1)^n}{\sqrt n}| \le \sum |\frac {(-1)^n}{\sqrt n}|$$
$$\sum |\frac {(-1)^n}{\sqrt n}| = \sum \frac 1{\sqrt n}$$
let $V_n=\frac 1{\sqrt n}=\frac 1{n^{1/2}}$
, $V_n$ is a Riemann serie that converge because $\alpha = 1/2 \lt 1$

Then we have $|\sum U_n| \le \sum V_n$ 
so $|\sum U_n|$ converge also, and then $\sum U_n$ is absolutely convergente ==> $\sum U_n$ converge
please i want to know if what im doing is correct ?
Thank u
 A: You are actually applying the criterion for Riemann series in the wrong way.

For $\alpha>0$, the series $\sum\limits_n n^{-\alpha}$ converges if and only if $\alpha>1$.

Therefore, the given series is not absolutely convergent.
However, since $x\mapsto x^{-1/2}$ is positive, decreasing and converging towards $0$ when $x$ goes to $+\infty$, using the alternating series criterion, the given series is convergent.
A: This is just an addendum to the previous answers, clearly showing that $-\eta\left(\frac{1}{2}\right)=\sum_{n\geq 1}\frac{(-1)^n}{\sqrt{n}}$ is a convergent series, but not an absolutely convergent series. My contribute is about computing an approximation for such series through the (inverse) Laplace transform. We have
$$ \frac{1}{\sqrt{n}}=\int_{0}^{+\infty}\frac{e^{-ns}}{\sqrt{\pi s}}\,ds \tag{1}$$
that can be stated as $\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)=\frac{1}{\sqrt{\pi s}}$. In particular:
$$ \sum_{n\geq 1}\frac{(-1)^n}{\sqrt{n}}=\int_{0}^{+\infty}\frac{1}{\sqrt{\pi s}}\sum_{n\geq 1}(-1)^n e^{-ns}\,ds = -\int_{0}^{+\infty}\frac{ds}{\sqrt{\pi s}(e^s+1)}\,ds \tag{2}$$
and the original series with alternating signs is converted into the integral of a smooth, decreasing function. With a change of variable ($s=t^2$) we also get:
$$ S=\sum_{n\geq 1}\frac{(-1)^n}{\sqrt{n}}=-\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{e^{t^2}+1} \tag{3}$$
hence it is really trivial that $S<0$, but we also get:
$$ \left| S\right| \leq \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{2+t^2+\frac{t^4}{2}} = \sqrt{\frac{\pi}{6}}.\tag{4} $$
A: If $a_n$ is a decreasing sequence of positive reals that goes to $0$ then $\sum\limits_{i=1}^\infty (-1)^ia_i$ converges
Proof: notice that


*

*$a_{2m}-a_{2m-1}\leq\sum\limits_{i=1}^{2m}(-1)^i a_i \leq a_{2m}$

*$-a_{2m+1}\leq\sum\limits_{i=1}^{2m+1}(-1)^i a_i \leq -a_{2m+1}+a_{2m}$
So by the sandwich theorem both the sequence of odd and even partial sums go to $0$.
A: No, you have missed the use of the Reimann series, it actually says that $\sum\frac1{\sqrt{n}}$ diverges.
So that tells you that the series does not converge absolutely.
Another way to see this is to compare the sum to the rectangles approximation to $\int_2^\infty n^{-1/2}$ which diverges.
Conditional convergence is tougher:  To lowest non-cancelling order in $n$,
$$
\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}} = \frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n^2+n}
=\frac{(n+1)\sqrt{n}-n\sqrt{n}\sqrt{1+\frac1n}}{n^2+n}
\\=\frac{n\sqrt{n}+\sqrt{n} -n\sqrt{n} \left(\sqrt{1+\frac1n}-1\right)-n\sqrt{n}}{n^2+n}
\approx \frac{n\sqrt{n}+\sqrt{n} -n\sqrt{n} \left(1+\frac1{2n}-1\right)-n\sqrt{n}}{n^2+n} 
\\=\frac{\sqrt{n} -n\sqrt{n} \left(\frac1{2n}\right)}{n^2+n}=
\frac{\sqrt{n} -\frac12\sqrt{n} }{n^2+n}=\frac12 \frac{\sqrt{n} }{n^2+n}<\frac12 \frac{\sqrt{n} }{n^2}= \frac12 \frac{1 }{n^{3/2}}
$$
which converges, so the series converges conditionally.
