Intuition behind the Divergence of series We know that the series is $\sum_{n=1}^ \infty \frac{1}{n}$ diverges.
But when we think intuitively, the sum of the series will grow very slowly after some stage, then how can we say that it diverges. How the series $\sum_{n=1}^ \infty \frac{1}{n^2}$ is different from the above.
Thank You.
 A: I think the Integral Test could help us to find the difference between these two series. According to this criteria, if $f(x)$ is positive, continuous and monotonic decreasing for $x\geq N$ and is such that $f(n)=u_n, n=N,N+1,...$, than $\sum u_n$ converges or diverges according as $$\int_N^{\infty}f(x)dx$$ converges or diverges. Now try to think of two functions $f(x)=1/x,~~~f(x)=1/x^2$. I hope to get the answer by yourself.
A: I hope your intuition tells you that the series $${1\over2}+{1\over4}+{1\over4}+{1\over8}+{1\over8}+{1\over8}+{1\over8}+{1\over16}+\cdots\tag1$$ diverges because, when you group like terms, it's just adding $1/2$ over and over. Then I hope your intuition tells you that $${1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+\cdots$$ grows faster than (1) since each term in it is at least as big as the corresponding term in (1) so it, too, must diverge. 
A: Perhaps intuition may come from  the comparison and related integral tests.
The comparison test shows that the first series diverges, as you can group the terms into subgroups that sum to greater than $\frac{1}{2}$, hence the sum can be made as large as you like. Explicitly, we can write the first $2^n$ terms as $\sum_{n=1}^{2^n} \frac{1}{n} = 1+\sum_{k=0}^n\sum_{l=0}^{2^{k-1}-1} \frac{1}{2^k+l}$, and it is straightforward to see that $\sum_{k=0}^n\sum_{l=0}^{2^{k-1}-1} \frac{1}{2^k+l} \geq \sum_{k=0}^n\sum_{l=0}^{2^{k-1}-1} \frac{1}{2^k} = \sum_{k=0}^n \frac{1}{2}=\frac{n+1}{2}$. Hence the series diverges.
If we let $f(x)= \frac{1}{x^2}$, we know that $\sum_{n=2}^N \frac{1}{n^2} = \sum_{n=2}^N f(n) \leq \int_1^{N-1} f(x) dx = 1-\frac{1}{N-1}$, and hence the series is bounded above, hence it converges.
A: To extend the Gerry Myerson's answer:
There is a very intuitive theorem which says that a series $\sum a_n$ with $a_n$ positive and (eventually) decreasing converges if and only if $\sum 2^n a_{2^n}$ converges.
Using it, one can turn $\sum \frac{1}{n}$ into $\sum 2^n\frac{1}{2^n}=\sum 1$ which diverges; and $\sum \frac{1}{n^2}$ into $\sum 2^n\frac{1}{2^{2n}}=\sum 2^{-n}$ which converges.
It is extremely handy with series like $\sum \frac{\log\log n}{n\log^2 n}$ &c.
