<s>[INST] Why does the $ p $-series diverge when $ p = 1 $? We’re currently working with $ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where $ p = 1 $, should converge, whereas in reality, it diverges. Can anyone demonstrate or link me to a proof or explanation of why this is so?
 [/INST] To answer your question, the integral test says that if $a_n$ is a positive decreasing sequence, and there exists a decreasing positive $f$ such that $f(n)=a_n$ for each $n$, then $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\int_1^\infty f(x) dx$$ does. You have correctly found that with $f(x)=\frac 1 x $ and $a_n=\frac 1 n$ we can prove divergence. Using the integral test, you can also show that any $p$ series converges for $p>1$ and diverges for $p\leq 1$.
Another elegant proof is the following. The $1$-series (a.k.a. Harmonic series), goes as follows. Let 
$$H_{n}=\sum_{k=1}^n \frac 1 k $$
Then $$\eqalign{
  & {H_2} = \sum\limits_{k = 1}^2 {{1 \over k}}  = 1 + 1 \cdot {1 \over 2}  \cr 
  & {H_{{2^2}}} = \sum\limits_{k = 1}^4 {{1 \over k}}  = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} > 1 + {1 \over 2} + {2 \over 4} = 1 + 2 \cdot {1 \over 2}  \cr 
  & {H_{{2^3}}} = \sum\limits_{k = 1}^8 {{1 \over k}}  = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} = 1 + 3 \cdot {1 \over 2}  \cr 
  & {H_{{2^4}}} = \sum\limits_{k = 1}^{16} {{1 \over k}}  = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} +  \cdots  + {1 \over {16}} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} + {8 \over {16}} = 1 + 4 \cdot {1 \over 2} \cr} $$
In general, we then show $${H_{{2^n}}} > 1 + {n \over 2}$$ which proves the Harmonic or $1$-series must diverge.
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