Does $\sum\limits_{n=1}^\infty \frac{1}{n^p}, 1 < p < 2$ converge? I know $\sum\limits_{n=1}^\infty \frac{1}{n^p}$ diverges when $p \le 1$ and converges when $p \ge 2$. How about $1 < p < 2$?
 A: By the Integral Test (note that the terms are monotone decreasing and positive), we may consider the integral
$$\int_1 ^ \infty \frac{1}{x^p} dx $$
Since $p > 1$, i.e. $p - 1 > 0$, we can evaluate the integral as
$$\left[\frac{1}{(1-p)x^{p-1}}\right]_1 ^ \infty = - \frac{1}{1-p} < \infty$$
A: The $p$-series converge for any $p > 1$. Observe that
$$ \sum_{n=1}^\infty 2^n \frac{1}{2^{np}} = \sum_{n=1}^\infty 2^{(1-p)n}
$$
By the Cauchy condensation test, $\sum_{n=1}^\infty n^{-p}$ is bounded by the above sum, and hence coverges.
A: Here's another cool test, the Cauchy Condensation Test: 
Let  $(a_n)$ be a  nonincreasing sequence of positive numbers. Then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=1}^\infty 2^n a_{2^n}$ converges. 
This is essentially a discrete change of variables (or substitution) formula, and as change of variables in integrals is a powerful tool, so is this test. The proof is very simple and follows from the monoonicity of the sequence. For example: 
Harmonic series $a_n =n^{-p}$. Then $a_{2^n} =2^{-np}$, and 
$$\sum_{n=1}^\infty 2^n a_{2^n} = \sum_{n=1}^\infty 2^{(1-p)n},$$ 
a geometric series, which converges if and only if $p>1$. 
Another example. $a_n = 1/(n \ln^\alpha  n)$. Then 
$a_{2^n} = \frac{1}{2^n n^{\alpha} (\ln 2)^{\alpha}}$. Therefore 
$\sum_{n=2}^{\infty}2^n a_{2^n} = \frac{1}{(\ln 2)^{\alpha}} \sum_{n=1}^\infty n^{-\alpha}$.
By last example this converges if and only if $\alpha>1$. 
You can continue and do more and more complex examples. Try 
$a_n = \frac{1} {n (\ln n) (\ln\ln n ) \dots (\ln \ln \ln \dots \ln n)^{\beta}}$, when the number of iterations of $\ln $ is fixed and equal to $k$.  
A: You already have two standard ways. Here is another, using comparison:
For $p>1$, we have
$$
\lim_{n\to+\infty}\frac{\frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}}{\frac{1}{n^p}}= p-1\neq 0.
$$
Now, the positive series
$$
\sum_{n=1}^{+\infty} \Bigl(\frac{1}{n^{p-1}}-\frac{1}{(n+1)^{p-1}}\Bigr)
$$
converges (it is a telescoping series with terms tending to $0$, the sum is actually equal to $1$). Hence, by comparison
$$
\sum_{n=1}^{+\infty}\frac{1}{n^p}
$$
also converges.
