# Question about series convergence $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$

So I have been playing around with convergent series recently and I still have a hard time understanding why $\sum_{n=1}^\infty \frac{1}{n}$ diverges and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.

I can write out the first couple of terms of $\sum_{n=1}^\infty \frac{1}{n}$ and compare them to a second series:

$\color{red}{1}+\color{green}{\frac{1}{2}}+\color{blue}{\frac{1}{3}+\frac{1}{4}}+\color{maroon}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+...$

$\color{red}{1}+\color{green}{\frac{1}{2}}+\color{blue}{\frac{1}{4}+\frac{1}{4}}+\color{maroon}{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}+...$

The colored terms in the second series always add up to $\frac{1}{2}$ and the corresponding colored terms in the first series always add up to more than $\frac{1}{2}$. Since the second series diverges the first one must also diverge. This still makes perfect sense. However if I write out the first couple of therms of $\sum_{n=1}^\infty \frac{1}{n^2}$

$\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}...$

then how does this converge?

Why is it not possible to group them similar to the first series? It seems to me that if I have infinitely many numbers I will be able to group $n$ of them to get to some number $k$ that I can add up forever.

• They decrease too fast in the series in question. Eventually, for no matter what real number you pick, you will never reach that number by adding the last terms. – Eoin Jan 24 '15 at 17:52
• @SimonS My mistake. I fixed it. – qmd Jan 24 '15 at 17:54
• Have you tried doing the same with the $\sum_{n=1}^\infty \frac{1}{n^2}$, like what you did with $\sum_{n=1}^\infty \frac{1}{n}$? It's the Cauchy condensation test, you can show $\sum_{n=1}^\infty \frac{1}{n^2}$ converges. – sciona Jan 24 '15 at 17:57

$$\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots$$

You want to walk a mile.

First you walk half of it. Then take a bit of rest. Then walk half of the remaining, then take a bit of rest. And so on.

Will you ever go beyond the mile?

You have an infinite number of numbers (of the form $\frac{1}{2^n}$), but you can never go beyond the mile. Basically, the numbers are decreasing too fast, that no matter how many you take, you still are bounded.

In this case, you can prove that (using $\frac{1}{n^2} \lt \frac{1}{(n-1)(n+1)}$)

$$1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} \lt 2 -\frac{1}{n}$$

So it is in fact bounded, and you cannot go beyond $2$, no matter how many terms you take.

(And any bounded monotonic sequence converges).

• Now I can actually visualize it. Thanks – qmd Jan 24 '15 at 18:18
• @Rzeta: You are welcome! – pedant Jan 24 '15 at 18:21

An easy way to see that $\sum \frac{1}{n^{2}}$ converges is to compare with the larger telescoping series $$1 + \sum_{n=2}^{\infty} \frac{1}{n(n-1)} = 1 + \sum_{n=2}^{\infty} \left[\frac{1}{n - 1} - \frac{1}{n}\right] = 2.$$

This ad hoc trick may be unsatisfying for a couple of reasons:

• How does one come up with similar estimates for other series? (Answer: One doesn't, at least not in an algorithmic way.)

• If someone hands you a general infinite series, how do you tell whether or not it converges? (Answer: That's difficult.)

But at least this estimate shows $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ converges, and the sum is between $1$ and $2$.