Two infinite series: the sum of reciprocals of natural numbers and their squares

两道无穷级数：自然数及其平方的倒数和 苏剑林 · 2009-08-05 · Mathematics series natural number reciprocal square

Prove whether the following series diverge or converge:

(1) $\sum_{x = 1}^\infty \frac{1}{x} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...$

(2) $\sum_{x = 1}^\infty \frac{1}{x^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...$

At first glance, since $1/x,1/{x^2}$ both tend to zero, one would expect these series to converge. But is that really the case? more

The result turns out to be surprising: series (1) diverges, while series (2) converges.

Proof:

(1)

$$\begin{aligned}1+1/2+1/3+1/4+1/5+... \\ =1+(1/2+...+1/{10})+(1/{11}+...+1/{100})+(1/{101}+...+1/{1000})+... \\ >1+1/{10}\cdot 9+1/{100}\cdot 90+1/{1000}\cdot 900+... \\ =1+9/{10}+9/{10}+9/{10}+...\end{aligned}$$

Since the terms above can be grouped indefinitely, there are infinitely many groups each summing to at least $9/{10}$, and adding them all up the total tends to infinity.

We could write out many similar proofs; for instance, Wikipedia gives an essentially identical argument: http://zh.wikipedia.org/wiki/%E8%B0%83%E5%92%8C%E7%BA%A7%E6%95%B0

(2)

$$\begin{aligned}1+1/{2^2}+1/{3^2}+1/{4^2}+... \\ =1+(1/{2^2}+...+1/{10^2})+(1/{11^2}+...+1/{100^2})+(1/{101^2}+...+1/{1000^2})+... \\ <1+1/{2^2}\cdot 9+1/{11^2}\cdot 90+1/{101^2}\cdot 900+... \\ <1+1/{2^2}\cdot 9+1/{10^2}\cdot 90+1/{100^2}\cdot 900+... \\ =4.24999...->4.25\end{aligned}$$

From this we can see that the value of this expression will never exceed 4.25.

The truth

In fact, series (1) is known as the "harmonic series", and as the number of terms tends to infinity, its value likewise tends to infinity.

$$\sum_{x = 1}^\infty \frac{1}{x} -> \infty$$

Series (2) is even more mysterious. In our proof above we used a loose bound, but in fact, exactly:

$$\sum_{x = 1}^\infty \frac{1}{x^2} = {\pi^2}/{6}$$

Mathematics is often full of surprises — who would have thought that the sum of the reciprocals of all square numbers turns out to be related to $\pi$!