Proof $\sum \frac{1}{n^a}$ is convergent for a > 1 I get to the fact that $\sum_{k=1}^n \frac{1}{k^a}$ < $\frac{1}{a-1} - \frac{1}{(a-1)(n+1)^{a-1}} - \frac{1}{(n+1)^a} + 1$ and hence $\sum_{k=1}^n \frac{1}{k^a}$ is bounded. How to deduce $\sum \frac{1}{n^a}$ is convergent ?
Note that a > 1
 A: Recall that a series of positive terms converges if the sequence of partial sums is bounded. Consider the $n$th partial sum of the series, 
$$s_n = \sum_{k = 1}^n \frac{1}{n^a}$$
For each $m \ge 0$, $\frac{1}{k^a} \le \frac{1}{2^{ma}}$ for $k = 2^m, 2^m+1,\ldots, 2^{m+1}-1$. Choose a natural number $M$ such that $2^{M+1} > n$. Since each integer $k \ge 1$ lies between $2^m$ and $2^{m+1}-1$ for a unique $m\ge 0$,
$$s_n \le  \sum_{m = 0}^M \sum_{2^m \le k < 2^{m+1}} \frac{1}{k^a} \le \sum_{m = 0}^M \sum_{2^m \le k < 2^{m+1}} \frac{1}{2^{ma}} = \sum_{m = 0}^M \frac{1}{2^{(a-1)m}}\le \sum_{m = 0}^\infty \left(\frac{1}{2^{a-1}}\right)^m$$
The last series is geometric with common ratio $\frac{1}{2^{a-1}}$; since $a > 1$, $\frac{1}{2^{a-1}} < 1$ so the geometric series converges (to $\frac{2^{a-1}}{2^{a-1}-1}$). Since $n$ was arbitrary, $(s_n)$ is bounded. Thus, the series $\sum\limits_{n = 1}^\infty \frac{1}{n^a}$ converges.
A: First of all, if $n < \infty$, then it is convergent no matter what. So I will argue for the case $n=\infty$.
There is an inequality, if $f$ is decreasing, then
$$ \sum_{k=2}^n f(k) \leq \int_1^n f(t)dt \leq \sum_{k=1}^{n-1} f(k)
$$
so rearranging a bit you will get upper and lower bounds for the integral or the sum. Thus the sum is convergent iff the integral is, so in this case
$$ \sum_{k=1}^\infty \frac{1}{k^\alpha} \leq 1 + \int_1^\infty \frac{dt}{t^{\alpha}}
$$
and I will let you sort out the details. Notice that it is only convergent for $\alpha > 1$, why? 
A: Using Cauchy Condensation test, the convergence of $\sum_{n=1}^\infty n^{-a}$ is equivalent to
$$ \sum_{n=1}^\infty 2^n \frac{1}{(2^n)^a} = \sum_{n=1}^\infty 2^{n(1-a)}
$$
The right side is a geometric series, and converges if and only if...
