# 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.

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It grows slower? I'm not sure what you're looking for here. – Alex Becker Feb 1 '13 at 10:41
What happens here is this kind of situation: $x_n \to +\infty$ but $x_{n+1}-x_n \to 0$. – Siméon Feb 1 '13 at 10:43
Neither of your summations should start at $n=0$. – Marc van Leeuwen Feb 1 '13 at 10:49
Suppose if we take the sum up to say 1000 terms. After that the sum will not grow rapidly. It grows with much slower rate. similarly with the case $\sum_{n=1}^ \infty \frac{1}{n^2}$. – Kumar Feb 1 '13 at 11:04
At $n=1000$ the term ${1\over n}$ is a thousand times greater than the term ${1\over n^2}$, and this ratio becomes worse when $n$ increases $\ldots$ – Christian Blatter Feb 1 '13 at 11:13

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.

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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.

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Of course, we had a thread recently asking about the intuition behind $\int dx/x = \ln x$. – Alex Becker Feb 1 '13 at 11:31
@AlexBecker: I am not aware of that. Honestly, I answered the OP's last question. If the OP wanted to get the differences by finding which is slow or which is slower, I personally would prefer Gerry's clear answer. I hope, I didn't make Reedy's mind distracted. Thanks Alex. :) – Babak S. Feb 1 '13 at 11:37

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.

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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.

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