# Integral and series convergence intuition

I have this problem I ran into during my studies to the upcoming exam: I don't feel I have the intuition of whether a series or an integral converges or not. What are the things I should look for when looking at a closed form expression, and thinking whether it converges or not? It's not as intuitive to me as in sequences, which come up much easier usually.

Thanks in advance for any advice! Also, not sure if this question deserves the 'soft question' tag. Feel free to add it if you think it's needed.

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Are you aware of the concept of asymptotic equivalence? That's where I get most of my 'intuition'. – Git Gud Jul 1 '13 at 21:15
No, never heard of it. – ohad Jul 1 '13 at 21:16

Something that I find helps my intuition is to have ready examples of series that converge and diverge. After all the only way you can "see" if a series converges is if it "looks" like a convergent series.

Series that converge which you should memorize:

$$\sum_{n=1}^\infty \frac{1}{n^2} \qquad \text{(converges)}$$

$$\sum_{n=1}^\infty \frac{1}{n^p} \qquad \text{(converges if p>1)}$$

$$\sum_{n=0}^\infty ar^n \qquad \text{(converges if |r|<1)}$$

Series which diverge that you should memorize:

$$\sum_{n=1}^\infty \frac{1}{n} \qquad \text{(diverges)}$$

$$\sum_{n=2}^\infty \frac{1}{\ln(n)} \qquad \text{(diverges)}$$

Something else to keep in mind is the rate of growth of different funcitons. In the long run we have: $$\ln(n) \lt n \lt n^2 \lt 5^n \lt n!$$ which means that $$\frac{1}{\ln(n)} \gt \frac{1}{n} \gt \frac{1}{n^2} \gt \frac{1}{5^n} \gt \frac{1}{n!}.$$

If you don't already know it you should learn the limit comparison test. This is a free licence to your intuition. For instance when determining the convergence of the following series:

$$\sum_{n=1}^\infty \frac{5n^3-3n^2+177}{6n^4-n^2+1}$$

We can squint our eyes and see that the top is basically $5n^3$ and the bottom is $6n^4$ telling us that,

$$\frac{5n^3-3n^2+177}{6n^4-n^2+1} \sim \frac{5n^3}{6n^4} = \frac{5}{6n} \sim \frac{1}{n}$$

So our intuition tells us this series should diverge because its really just a harmonic series in disguise. We justify that intuition by applying the limit comparison test.

Another example:

$$\sum_{n=1}^\infty \frac{2^n}{n!}$$

We have a ratio of $2^n$ and $n!$. Recall that $n!$ grows much faster than $2^n$, this means its contribution is much more significant than the contribution of $2^n$. On its own the $n!$ in the denominator would make the series converge therefore we can guess that this series will still converge. We verify our guess by applying the ratio test (this should be your go-to convergence test when factorials are involved).

$$\left\vert \frac{a_{n+1}}{ a_n} \right\vert = \frac{2^{n+1}(n)!}{(n+1)!2^n} = \frac{2}{(n+1)} \rightarrow 0 \text{ as } n \rightarrow \infty$$

Since the ratio of consecutive terms approaches $0$, which is less than $1$, as $n$ approaches $\infty$ we can conclude that the series converges.

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Very nice answer. – Potato Jul 1 '13 at 22:37
Hi, thank you! Very useful! – ohad Jul 2 '13 at 5:37

It's not as intuitive to me as in sequences, which come up much easier usually.

Sums/Integrals are sequences, they have the same properties as sequences, to see this consider a sum/integral as partial sums/integrals, now the value of each partial sum/integral is just a term of sequence, if you can see the sequence converges/diverges then the sum/integral converges/diverges ( by the very definition of sum/integral).

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