We basically have to prove that,\[\sum_{n=2}^{n}\dfrac{1}{n^2}<1\]
\(\text{Motivation of proof}:\),the first idea that struck me was the use of telescoping series,if i could write,\(\dfrac{1}{n^2}<\text{something}\) and write,\(\dfrac{1}{(n-1)^2}<\text{another something}\),and when we add the terms of \(L.H.S\) we would get the required expression and when we add the terms of the \(R.H.S\) we would get a telescoping series.The next thing that came to my mind was that,\[n^2>(n)(n-1)\](we have taken the greater than sign as then when we take the reciprocal the sign would get reversed),since \(\dfrac{1}{(n)(n-1)}=\dfrac{1}{n-1}-\dfrac{1}{n}\),and we would get a telescoping series like this,
\[\dfrac{1}{2^2}<\dfrac{1}{1}-\dfrac{1}{2}\\
\dfrac{1}{3^2}<\dfrac{1}{2}-\dfrac{1}{3}\\
.\\
.\\
.\\
\dfrac{1}{n^2}<\dfrac{1}{n-1}-\dfrac{1}{n}\\
\text{adding,we get}:\\
\sum_{n=2}^{n}\dfrac{1}{n^2}<1-\dfrac{1}{n}<1\].Hence proved.And done!

@Calvin Lin
–
@Calvin Lin Oops, although Induction is tempting at first look, it isn't the best way to go about. I haven't found an inductive proof yet 😕.

from the basel problem we get \[\sum_{n=1}^\infty (\frac{1}{n^2}) = \frac{\pi^2}{6}<2\]. although this is probably not intended, but still works as a good proof. we prove the basel proble:\[\]
Euler's original derivation of the value \(\frac{π^2}{6}\) essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series.

recall the Taylor series expansion of the sine function

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots. \]
Dividing through by x, we have

\[\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots. \]
Using the Weierstrass factorization theorem, it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials

\[\begin{align}
\frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\
&= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots.
\end{align}\]
If we formally multiply out this product and collect all the x2 terms (we are allowed to do so because of Newton's identities), we see that the x2 coefficient of sin(x)/x is

\[ -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) =
-\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.\]
But from the original infinite series expansion of sin(x)/x, the coefficient of x2 is −1/(3!) = −1/6. These two coefficients must be equal; thus,

\[-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.\]
Multiplying through both sides of this equation by \(-\pi^2\) gives the sum of the reciprocals of the positive square integers.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestWe basically have to prove that,\[\sum_{n=2}^{n}\dfrac{1}{n^2}<1\] \(\text{Motivation of proof}:\),the first idea that struck me was the use of telescoping series,if i could write,\(\dfrac{1}{n^2}<\text{something}\) and write,\(\dfrac{1}{(n-1)^2}<\text{another something}\),and when we add the terms of \(L.H.S\) we would get the required expression and when we add the terms of the \(R.H.S\) we would get a telescoping series.The next thing that came to my mind was that,\[n^2>(n)(n-1)\](we have taken the greater than sign as then when we take the reciprocal the sign would get reversed),since \(\dfrac{1}{(n)(n-1)}=\dfrac{1}{n-1}-\dfrac{1}{n}\),and we would get a telescoping series like this, \[\dfrac{1}{2^2}<\dfrac{1}{1}-\dfrac{1}{2}\\ \dfrac{1}{3^2}<\dfrac{1}{2}-\dfrac{1}{3}\\ .\\ .\\ .\\ \dfrac{1}{n^2}<\dfrac{1}{n-1}-\dfrac{1}{n}\\ \text{adding,we get}:\\ \sum_{n=2}^{n}\dfrac{1}{n^2}<1-\dfrac{1}{n}<1\].Hence proved.And done!

Log in to reply

I like your motivations haha :P

Log in to reply

From where you got this motivation ?

Log in to reply

I am sorry but i don't understand the meaning of your comment.Could you please explain?

Log in to reply

Simplification: Source of the question = ?Log in to reply

Log in to reply

Well, yours is the simplest way. For the sake of mentioning, one can also write a proof by induction for this.

Log in to reply

How do you do a proof by induction (that is fundamentally different from his approach)?

Log in to reply

@Calvin Lin Oops, although Induction is tempting at first look, it isn't the best way to go about. I haven't found an inductive proof yet 😕.

Log in to reply

\[ \sum_{i=1}^n \frac{1}{i^2 } < 2 - \frac{1}{n}. \]

This is similar to what Adarsh did.

Log in to reply

from the basel problem we get \[\sum_{n=1}^\infty (\frac{1}{n^2}) = \frac{\pi^2}{6}<2\]. although this is probably not intended, but still works as a good proof. we prove the basel proble:\[\] Euler's original derivation of the value \(\frac{π^2}{6}\) essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series.

recall the Taylor series expansion of the sine function

\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots. \] Dividing through by x, we have

\[\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots. \] Using the Weierstrass factorization theorem, it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials

\[\begin{align} \frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ &= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align}\] If we formally multiply out this product and collect all the x2 terms (we are allowed to do so because of Newton's identities), we see that the x2 coefficient of sin(x)/x is

\[ -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.\] But from the original infinite series expansion of sin(x)/x, the coefficient of x2 is −1/(3!) = −1/6. These two coefficients must be equal; thus,

\[-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.\] Multiplying through both sides of this equation by \(-\pi^2\) gives the sum of the reciprocals of the positive square integers.

\[\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.\]

Log in to reply

You are right,it wasn't intended but if you write this,you are expected to prove it too.

Log in to reply

fine

Log in to reply

Thanks a lot for adding the proof!

Log in to reply

Isn't this reimann zeta(2)? Then the value will be 1.64 only.

Log in to reply

Yes,it is but that isn't the intended proof,or else you will have to prove how you found the value of \(\zeta{(2)}\).

Log in to reply

Oh so that's why proof! Btw my method was just like yours

Log in to reply