<s>[INST] Why ${ \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent , but ${ \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent? I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit is equal to ${ \displaystyle\frac{\pi^2}{6} }$. In both cases, ${n^{th}}$ term tends to zero, so what makes these series different?
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} +  }$ ...
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} =1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + }$ ...
 [/INST] The following example may be easier to grasp. Consider the series
$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots. $$
So we have $2$ copies of $\frac{1}{2}$, $4$ copies of $\frac{1}{4}$, $8$ copies of $\frac{1}{8}$, $16$ copies of $\frac{1}{16}$, and so on like that forever. 
The $2^k$ copies of $\frac{1}{2^k}$ add up to $1$, so partial sums get arbitrarily large, and therefore our series diverges. 
Now look at the series of the squares of the above numbers. We get
$$1+\frac{1}{4}+\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\cdots $$
The first entry is $1$. The next $2$ entries add up to $\frac{1}{2}$. The next $4$ entries add up to $\frac{1}{4}$. The next $8$ entries add up to $\frac{1}{8}$. And so on forever. So the full sum is 
$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$ This geometric series has sum $2$.
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