Why is a divergent/convergent series multiplied with a constant a divergent/convergent series again? I was trying really hard to find a series smaller than  $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$ to prove, that $\sum\limits_{k=1}^{\infty} \frac{1}{2k}$ is divergent. Now I got to know that I can show that by using the harmonic series, because $0.5 \sum\limits_{k=1}^{\infty} \frac{1}{k} $ is divergent as well. I can't really understand that since $\frac{1}{2k}$ is clearly smaller than $ \frac{1}{k}$?
Furthermore, does this work for convergent series as well? Take the series $\sum\limits_{k=1}^{\infty } {(-1)^k} \frac{-1}{k^{2} } $ for example. Can I say, that series has to be convergent since $ \sum\limits_{k=1}^{\infty } {(-1)^k} \frac{1}{k^{2} }$ is convergent (because there, I can use Leibniz) ? 
Thanks for helping.
 A: We can easily show the result by the definition of limit for sequences.
Indeed, for example for the harmonic series divergent case, let $S_n=\sum_{k=1}^{n} \frac{1}{k}$ and we know that $S_n \to \infty$ which means that
$$\forall M_0\in \mathbb{R} \quad \exists \bar n\in \mathbb{N} \quad \forall n\ge \bar n \quad S_n\ge M_0$$
which means that we can make $S_n$ larger than any fixed number $M$.
Then consider $R_n=\frac12 \cdot S_n=\sum_{k=1}^{n} \frac{1}{2k}$ and we have that
$$\forall M=\frac12M_0\in \mathbb{R} \quad \exists \bar n\in \mathbb{N} \quad \forall n\ge \bar n \quad R_n=\frac12\cdot S_n\ge\frac 12 \cdot M_0= M$$
that is $R_n \to \infty$.
The same argument applies fro any divergent series and for any constant and a similar argument applies for convegent series.
For the series $\sum\limits_{k=1}^{\infty } {(-1)^k} \frac{1}{k^{2} }$ we don't need to use Leibniz since it converges absolutely that is
$$\sum\limits_{k=1}^{\infty } \left|{(-1)^k} \frac{1}{k^{2} }\right|=\sum\limits_{k=1}^{\infty } \frac{1}{k^{2} }<\infty \implies \sum\limits_{k=1}^{\infty } {(-1)^k} \frac{1}{k^{2} }<\infty$$
We are forced to use Leibniz, for example, to show the convergence of the series $\sum\limits_{k=1}^{\infty } {(-1)^k} \frac{1}{k }$.
A: Suppose that $\sum_{k= 1}^\infty\frac1{2k}=L$ for some $L\in\Bbb R$. Then, by the definition of convergence of a series, we will had that for any chosen $\epsilon>0$ there is a $N\in\Bbb N$ such that
$$\left|\sum_{k=1}^m\frac1{2k}-L\right|=\frac12\left|\sum_{k=1}^m\frac1k-2L\right|<\epsilon
\iff\left|\sum_{k=1}^m\frac1k-2L\right|<2\epsilon$$
for all $m\ge N$. However we knows that the last inequality is not true because the series $\sum_{k=1}^\infty\frac1k$ doesn't converge to a value of $\Bbb R$, and by assumption $2L\in\Bbb R$.
A: You may use this inequality with a telescopic RHS derived by Padé approximants:
$$ \frac{1}{2k}>\frac{1}{4}\log\left(\frac{1+3k+3k^2}{1-3k+3k^2}\right) \tag{1}$$
to immediately deduce that
$$ \frac{H_n}{2}=\sum_{k=1}^{n}\frac{1}{2k} > \frac{1}{4}\log(3n^2+3n+1).\tag{2} $$
$(2)$ also leads to $\gamma>\frac{1}{2}\log(3)$, not bad.
A: A simple solution:
$\sum\limits_{k=1}^\infty \frac{1}{2k}=\sum\limits_{k=1}^\infty \int\limits_0^1t^{2k-1} dt=\int\limits_0^1 \frac{1}{t} \sum\limits_{k=1}^\infty(t^2)^k dt=\int\limits_0^1 \frac{t}{1-t^2} dt=\big[\ln\frac{1}{\sqrt{(1-t^2)}}\big]_0^1=\infty$
