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} + \dots
(2) \sum_{x = 1}^\infty \frac{1}{x^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots
At first glance, since 1/x and 1/x^2 both tend to zero, they might seem to be convergent. Is this really the case?
The results are surprising: Series (1) diverges, and series (2) converges.
Proof:
(1)
\begin{aligned} & 1+1/2+1/3+1/4+1/5+\dots \\ & =1+(1/2+\dots+1/10)+(1/11+\dots+1/100)+(1/101+\dots+1/1000)+\dots \\ & >1+\frac{1}{10}\cdot 9+\frac{1}{100}\cdot 90+\frac{1}{1000}\cdot 900+\dots \\ & =1+9/10+9/10+9/10+\dots \end{aligned}
Since the terms can be grouped infinitely in this manner, there are infinitely many 9/10 terms being added; thus, the sum tends to infinity.
Many similar proofs can be written; for instance, Wikipedia provides a very similar proof: https://en.wikipedia.org/wiki/Harmonic_series
(2)
\begin{aligned} & 1+1/2^2+1/3^2+1/4^2+\dots \\ & =1+(1/2^2+\dots+1/10^2)+(1/11^2+\dots+1/100^2)+(1/101^2+\dots+1/1000^2)+\dots \\ & <1+\frac{1}{2^2}\cdot 9+\frac{1}{11^2}\cdot 90+\frac{1}{101^2}\cdot 900+\dots \\ & <1+\frac{1}{2^2}\cdot 9+\frac{1}{10^2}\cdot 90+\frac{1}{100^2}\cdot 900+\dots \\ & =4.24999\dots \to 4.25 \end{aligned}
From this, it can be seen that the value of this expression will never exceed 4.25.
The Truth
In fact, series (1) is known as the “Harmonic Series.” As the number of terms tends to infinity, its value also tends to infinity. \sum_{x = 1}^\infty \frac{1}{x} \to \infty Series (2) is even more mysterious. While we used an upper bound in our proof, the exact value is actually: \sum_{x = 1}^\infty \frac{1}{x^2} = \frac{\pi^2}{6} Mathematics is often surprising; the sum of the reciprocals of all square numbers is actually related to \pi!
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