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.

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## Comments

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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!

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I like your motivations haha :P

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From where you got this motivation ?

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

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

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

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

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@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 😕.

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

This is similar to what Adarsh did.

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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.

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

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Thanks a lot for adding the proof!

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

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fine

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

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

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

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