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I am trying to interpret an equation, but can't understand how the less-than signs work:

$\sum _{n=1}^{\infty } \frac{1}{n^2}<1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}=1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)=1+1=2$

The first part returns False when I put it into Mathematica which I don't think is how it should be interpreted since this value approximates 1.644.

$\sum _{n=1}^{\infty } \frac{1}{n^2}<1$

Squan Schmaan
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  • $$\sum_{n=1}^\infty\frac1{n^2}<1+\sum_{n=1}^\infty\frac1{n(n+1)}$$ is true. $$\sum_{n=1}^\infty\frac1{n^2}<1$$ is false. – Angina Seng Apr 06 '20 at 12:06
  • Do you agree that $3<1+4$ is true? Nothing about that notation ($+$ happening before $<$) changes when there are summations around. – Mark S. Apr 06 '20 at 12:07
  • The order of operations is such that the inequality is evaluated last, much like an equality would. – Wouter Apr 06 '20 at 12:09
  • Thanks Mark, sorry for the dumb question. So it should return a logical answer then.. But how does it result in a numerical conclusion, i.e = 1 + 1 = 2? – Squan Schmaan Apr 06 '20 at 12:09
  • @DanielSmith The entire line is claimed to be logically true. "$3<1+4=5$" could be a shorthand for "it's true that $3<1+4$ and it's true that $1+4=5$" with an implied "note that $3<5$". A similar thing is happening here: Each step is true, and the notable conclusion is that the original summation is less than $2$. – Mark S. Apr 06 '20 at 12:15
  • @MarkS. This makes sense, but what is the point in including 3, or in my case the sum of 1/N^2 to infinity. To me, it doesn't really add anything to the final conclusion. – Squan Schmaan Apr 06 '20 at 23:12
  • @DanielSmith Wouter spelled things out better, but I think you have things backwards. I don't know the context in your book/class/source, but it's not obvious that the sum of $1/n^2$ is even finite (I mean, the sum of $1/n$ isn't), or if it is finite that it's not like a million. But the argument in your post is a quick way to show it's less than $2$, even though you don't know what the value is. – Mark S. Apr 07 '20 at 02:05

2 Answers2

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Your statement "unpacks"

First $$\sum _{n=1}^{\infty } \frac{1}{n^2}<1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}$$ second $$1+\sum _{n=1}^{\infty } \frac{1}{n (n+1)}=1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)$$ third $$1+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+1)}\right)=1+1$$ fourth $$1+1=2$$

The use of multiple equalities and inequalities like this is a bit abuse of notation. What if I wrote $$1+1+1=1+2=3$$ then you might be tempted to say $1+1+1=1+2$ is true, so the above becomes $$\text{true}=3$$ (and in certain programming languages the statement really would be interpreted like that) but this is almost never what is meant. What is meant is that the series of (in)equalities should be unpacked, here as $1+1+1=1+2$ and $1+2=3$, independently.

Wouter
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Since $ \left(\forall n\geq 2\right),\ n^{2}\geq n\left(n-1\right) $, we get that $ \left(\forall n\geq 2\right),\ \frac{1}{n^{2}}\leq\frac{1}{n\left(n-1\right)} $, thus : $$ \sum_{n=1}^{+\infty}{\frac{1}{n^{2}}}=1+\sum_{n=2}^{+\infty}{\frac{1}{n^{2}}}<1+\sum_{n=2}^{+\infty}{\frac{1}{n\left(n-1\right)}}=1+\sum_{n=1}^{+\infty}{\frac{1}{n\left(n+1\right)}} $$

I suppose you can now understand why the first inequality is true.

CHAMSI
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