Why does the harmonic series diverge but the p-harmonic series converge I am struggling understanding intuitively why the harmonic series diverges but the p-harmonic series converges. I know there are methods and applications to prove convergence, but I am only having trouble understanding intuitively why it is. I know I must never trust my intuition, but this is hard for me to grasp. In both cases, the terms of the series are getting smaller, hence are approaching zero, but they both result in different answers.
$$\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \text{diverges}$$   $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\text{converges}$$

We produce two series that are close in spirit to the series you mentioned. Perhaps the divergence of the first, and the convergence of the second, will be clearer.
Consider the series
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\cdots.$$
So there is $1$ term equal to $\frac{1}{2}$, then a block of $2$ each equal to $\frac{1}{4}$, then a block of $4$ each equal to $\frac{1}{8}$, then a block of $8$ each equal to $\frac{1}{16}$, and so on forever. Each block has sum $\frac{1}{2}$, so if you add enough terms, your sum will be very big. But it will take an awful lot of terms to add up to $1000$, many more more terms than there are atoms in the universe.  Note that each term is less than the corresponding term in the harmonic series, so if you add together enough terms of the harmonic series, the sum will be very big.
Now consider the series
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\cdots.$$
Each term is $\ge$ the corresponding term in the series $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\cdots$.
Again, we find the sums of the blocks. The first block has sum $1$. The second has sum $\frac{1}{2}$. The third has sum $\frac{1}{4}$. The fourth has sum $\frac{1}{8}$, and so on forever. So if we add up "all" the terms, we get sum $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$, a familiar series with sum $2$. 
