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If I solve it by a Riemann sum approach i.e. $$\int_{1}^{\infty} 1/x^2 dx$$ then I can see that the sum of the geometric series is equals to 1. He mentioned that it should not be equals to 1, but instead the inequality $\leq 1$. It made sense to me at first, but after that it seemed to be even more illogical.

The fact that the first term itself is 1, and you are adding on 1/4, 1/9 .... should make it strictly more than 1. So how can it be $\leq 1$?

Isaac Browne
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statsguy21
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  • Take a look at this page, especially the second image, to see what is going on. – Jair Taylor Aug 31 '18 at 01:12
  • I see...very interesting. So my inequality was indeed wrong then, it should have been $\leq2$ instead. A quick question though. I cannot just integrate the curve from 0 to infinity because there is a vertical asymptote (y-axis), right? That's why I have to add 1 to the definite integral of the curve instead. – statsguy21 Aug 31 '18 at 01:32
  • Yes, $\int_0^\infty 1/x^2 , dx = \infty$. – Jair Taylor Aug 31 '18 at 01:36
  • You can determine that $\sum_{k=1}^\infty \frac{1}{k^2} \leq 2$, yes, if that is what you mean. – Jair Taylor Aug 31 '18 at 01:37
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    Aside: this is not a "geometric series". A "geometric series" is one where successive terms have a constant common ratio: e.g. $\sum_{n=0}^{\infty} a r^n = a + ar + ar^2 + ar^3 + \ldots $ –  Aug 31 '18 at 02:06

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The integral $\int_{1}^{\infty} 1/x^2 dx$ does not equal the sum $\sum_{n = 1}^{\infty} 1/n^2$.

Evaluating the integral gives a convergence to the value 1. Think of this integral as taking the area under the curve $1/x^2$ from x=1 to x=$\infty$. This convergence can be observed through the p-test, where p is the power that x is raised to. If p > 1, then the series converges. The exact value of convergence can be calculated by taking the antiderative and plugging in the limits. $$\int_{1}^{\infty} 1/x^2 dx = -\frac{1}{x}\Big|_1^\infty = -\frac{1}{\infty}-(-1) = 0 +1 = 1$$

The sum $\sum_{n = 1}^{\infty} 1/n^2$ evaluates to $\frac{{\pi}^2}{6} \approx 1.644934$. This is what you are thinking of when you mention taking 1 as the first term, then adding 1/4, 1/9... This is the famous Basel problem that Euler solved in the 1700's. A simple proof (Euler's approach) can be found here.

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    And many more proofs here: https://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-k-1-infty-frac1k2-basel-pro – Hans Lundmark Aug 31 '18 at 06:58