I'm trying to prove by comparison the convergence of generalized harmonic series, of the form $$\sum_{n=1}^{\infty} \frac{1}{n^{a+1}}$$ where $a>0$. I proved this for $a=1$, showing that $$\sum_{k=1}^{n} \frac{1}{k^2}<1+\sum_{k=2}^{n} \frac{1}{k(k-1)}$$ but I'm not succeeding in generalize this for any $a>0. $
Is it possible to prove it along the way of the first?
For $a>1$ you have $$n^{a+1}\ge n^2$$
Thus $$\frac {1}{n^{1+a}} \le\frac {1}{n^2}$$
By comparison we have $$\sum _1^\infty \frac {1}{n^{1+a}}\le \sum _1^\infty \frac {1}{n^2} =\pi^2/6$$
For $a>0$ the integral test will do.
With the Cauchy condensation test:
$\sum_{n=1}^{\infty} \frac{1}{n^{a+1}}$ is convergent $ \iff \sum_{n=1}^{\infty} 2^n\frac{1}{(2^n)^{a+1}}$ is convergent.
Can you proceed ?
This is along the same lines as your approach using $\frac1{(k-1)k}=\frac1{k-1}-\frac1k$.
Preliminary Inequality
For $0\le a\le1$, we have
$$
\begin{align}
\frac1{(k-1)^a}-\frac1{k^a}
&=\frac{k^a-(k-1)^a}{((k-1)k)^a}\tag{1a}\\
&=\frac{1-\left(1-\frac1k\right)^a}{(k-1)^a}\tag{1b}\\
&\ge\frac{\frac{a}k}{(k-1)^a}\tag{1c}\\[3pt]
&\ge\frac{a}{k^{1+a}}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a):}$ combine fractions
$\text{(1b):}$ divide numerator and denominator by $k^a$
$\text{(1c):}$ Bernoulli's Inequality
$\text{(1d):}$ $k\gt k-1$
Thus, $$ \frac1{k^{1+a}}\le\frac1a\left(\frac1{(k-1)^a}-\frac1{k^a}\right)\tag2 $$
Application of the Inequality
$$
\begin{align}
\sum_{k=1}^\infty\frac1{k^{1+a}}
&=1+\sum_{k=2}^\infty\frac1{k^{1+a}}\tag{3a}\\
&\le1+\frac1a\sum_{k=2}^\infty\left(\frac1{(k-1)^a}-\frac1{k^a}\right)\tag{3b}\\[2pt]
&=1+\frac1a\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a):}$ pull out the $k=1$ term
$\text{(3b):}$ apply $(2)$
$\text{(3c):}$ Telescoping Series
Thus, the series converges for $a\gt0$.
Taking It To The Limit
In this answer it is shown that $\lim\limits_{a\to0}\left(\zeta(1+a)-\frac1a\right)=\gamma$, the Euler-Mascheroni Constant. This implies that $$ \lim_{a\to0}\left(1+\frac1a-\sum_{k=1}^\infty\frac1{k^{1+a}}\right)=1-\gamma\tag4 $$
