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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.

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    $\begingroup$ If you write down the definition of convergence (with $\epsilon$ and $N$) you should be able to see why. One set of partial sums is just a constant times the other set. $\endgroup$ – Ethan Bolker Dec 8 '18 at 23:27
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    $\begingroup$ The claim in the title is not completely true. The divergent series $\sum_k k$ can become convergent if you multiply it with the constant $0$. (But that is the only constant this works for). $\endgroup$ – Henning Makholm Dec 8 '18 at 23:47
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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 $2L\in\Bbb R$.

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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 $\sum\limits_{k=1}^{\infty } {(-1)^k} \frac{1}{k }$.

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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.

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A simple solution:

$\sum\limits_{k=1}^\infty \frac{1}{2k}=\sum\limits_{k=1}^\infty \int\limits_0^1t^{2k-1} dt=\int\limits_0^1 \frac{1}{t} \sum\limits_{k=1}^\infty(t^2)^k dt=\int\limits_0^1 \frac{t}{1-t^2} dt=\big[\ln\frac{1}{\sqrt{(1-t^2)}}\big]_0^1=\infty$

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