$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$. Why is this inequality true? The proof to the generalized harmonic series has the following inequality:
$$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$$
with $c > 1$ and $c = 1 + \omega$, $\omega > 0$.
I am aware that the generalized harmonic series can be proven alternatively. But I don't see the above inequality. Why is that true? I tried to list out a few of the first terms and compare them term-by-term.
LHS: $ 1 + \frac{1}{2\cdot2^{\omega}} + \frac{1}{3\cdot3^{\omega}} + \frac{1}{4\cdot4^{\omega}} + \frac{1}{5\cdot5^{\omega}} + \dots$
RHS: $ 1 + \{\frac{1}{2^{\omega}} + \frac{1}{2^{2\omega}} + \frac{1}{2^{3\omega}} + \frac{1}{2^{4\omega}} + \dots\} = 1 + \{\frac{1}{2^{\omega}} + \frac{1}{4^{\omega}} + \frac{1}{8^{\omega}} + \frac{1}{16^{\omega}} + \dots\}$
Starting from $k = 5$, however, the term of LHS is not necessarily smaller than that of RHS.

Group by packets of size $2^n$:
$$\sum_{k=1}^{+\infty}\frac 1 {k^c}=\sum_{n=0}^{+\infty}\left(\sum_{m=2^n}^{2^{n+1}-1}\frac 1 {m^c}\right)< \sum_{n=0}^{+\infty}\frac{2^n}{(2^n)^c}=\sum_{n=0}^{+\infty}\frac 1 {2^{\omega n}}$$
