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