summation

Could we do this by limit of a sum?

\lim_{n\to\infty}\sum_{i = 1}^{n}{\frac{1}{i^2}}

Okay sandi has got it lol.

ur answer to the sum of squares od reciprocals of odd terms...

see from above

11² + 12² + 13² + ...... = (pi)²6

also

11² - 12² + 13² -......... = (pi)²12 (i learnt it as a fact..)

so adding both series we get....

2(11² + 13² + ...) = 3 (pi)² 12

so 11² + 13² + ... = (pi)² 8

[1][1][1]

An alternative:
We start by proving two results, seemingly unrelated to the given problem:

1) \sum_{k=1}^n \cot^2\left(\frac{k\pi}{2n+1}\right) = \frac{n(2n-1)}{3}

2) \sum_{k=1}^n \csc^2\left(\frac{k\pi}{2n+1}\right) = \frac{2n(n+1)}{3}

The second one can be proved from the first one by using the identity \csc^2\varphi = \cot^2 \varphi + 1, so I prove only the first one.

Using De' Moivre's formula, we have
\cos n\varphi + i \sin n\varphi = (\cos\varphi + i \sin\varphi)^n

Expand the right-side binomially:
(\cos\varphi + i \sin\varphi)^n = \sum_{r=0}^n\, ^nC_r i^r \cos^{n-r}\varphi \ \sin^r\varphi
On the RHS, for each even value of r the summand is purely real, while for each odd value of r, it odd. Taking this fact into account and collecting the real and imaginary quantities, we obtain

\cos n \varphi + i \sin n\varphi = (\cos^n\varphi - \, ^n\!C_2 \cos^{n-2}\varphi \sin^2\varphi +\, ^n\!C_4 \cos^{n-4}\varphi \sin^4\varphi - \ldots)
+i\ (\,^nC_1 \cos^{n-1}\varphi \sin \varphi - \,^n\!C_3 \cos^{n-3}\varphi \sin^3\varphi +\, ^n\!C_5 \cos^{n-5}\varphi \sin^5\varphi - \ldots)

Equating the imaginary parts we obtain,
\sin n\varphi = \,^n\!C_1 \cos^{n-1}\varphi \sin \varphi - ^n\!C_3 \cos^{n-3}\varphi \sin^3\varphi +\, ^n\!C_5 \cos^{n-5}\varphi \sin^5\varphi - \ldots

Replacing n by 2n+1, write the above equality as

\sin(2n+1)\varphi =\, ^{2n+1}C_1 \cos^{2n}\varphi \sin\varphi -\, ^{2n+1}C_3 \cos^{2n-2}\varphi \sin^3\varphi +\, ^{2n+1}C_5 \cos^{2n-4}\varphi \sin^5\varphi + \ldots

\Rightarrow\sin(2n+1)\varphi =\sin^{2n+1}\varphi (\, ^{2n+1}C_1 \cot^{2n}\varphi -\, ^{2n+1}C_3 \cot^{2n-2}\varphi +\, ^{2n+1}C_5 \cot^{2n-4}\varphi + \ldots )\qquad (1)

Now, if we take
\varphi = \frac{k\pi}{2n+1},\, k =1,\,2,\,3,\,\ldots,\,n,
then for each of these values, the LHS of the above equation is zero. But since for these values of φ, sin φ is not zero, the quantity inside the bracket must be zero for these values of φ . That means the numbers
x_k = \cot^2\left(\frac{k\pi}{2n+1}\right),\, k =1,\,2,\,3,\,\ldots,\,n
must be the roots of the -th degree polynomial equation
^{2n+1}C_1 x^n -\, ^{2n+1}C_3 x^2 + \,^{2n+1}C_5 x^3 - \ldots =0
As such the sum of the roots of this equation
\boxed{\sum_{k=1}^n \cot^2\left(\frac{k\pi}{2n+1}\right) = \dfrac{^{2n+1}C_3}{^{2n+1}C_1}=\dfrac{n(2n-1)}{3}}

proving the proposition 1) above. As said already, 2) can be obtained from 1), so I do not prove the second one.

Next, it is easy to prove that for a non-zero acute angle (in radians), the following double inequality holds:
\sin\alpha < \alpha < \tan\alpha

(Can someone prove it ? )

From this, we obtain the inequalities
\cot^2\alpha < \left(\frac{1}{\alpha}\right)^2 < \csc^2\alpha\qquad (2)

Hence, we obtain the following sequence of double inequalities:

\cot^2\left(\frac{k\pi}{2n+1} \right)< \left(\frac{2n+1}{k\pi}\right)^2 < \csc^2\left(\frac{k\pi}{2n+1}\right),\,k=1,\,2,\,3,\,\ldots,\,n.
Summing them up and using 1) and 2) already derived, we get
\dfrac{n(2n-1)}{3} < \sum_{k=1}^n \left(\frac{2n+1}{k\pi}\right)^2 < \dfrac{2n(n+1)}{3}
From which we obtain
\dfrac{n(2n-1)}{3}\cdot \dfrac{\pi^2}{(2n+1)^2} < \sum_{k=1}^n \dfrac{1}{k^2} < \dfrac{2n(n+1)}{3}\cdot \dfrac{\pi^2}{(2n+1)^2}\qquad (3)
The left side is
\dfrac{2n}{2n+1}\cdot \dfrac{2n-1}{2n+1}\cdot \dfrac{\pi^2}{6}=\left(1-\dfrac{1}{2n+1}\right)\left(1-\dfrac{2}{2n+1}\right)\dfrac{\pi^2}{6}
while the right side is
\dfrac{2n}{2n+1}\cdot \dfrac{2n+2}{2n+1}\cdot \dfrac{\pi^2}{6}=\left(1-\dfrac{1}{2n+1}\right)\left(1+\dfrac{1}{2n+1}\right)\dfrac{\pi^2}{6}

Hence, from , we obtain the following inequality:
\boxed{\left(1-\dfrac{1}{2n+1}\right)\left(1-\dfrac{2}{2n+1}\right)\dfrac{\pi^2}{6} < \sum_{k=1}^n \dfrac{1}{k^2} < \left(1-\dfrac{1}{2n+1}\right)\left(1+\dfrac{1}{2n+1}\right)\dfrac{\pi^2}{6}}
Taking the limit as n→∞ and applying Squeeze theorem, we get:
\sum_{k=0}^\infty \dfrac{1}{k^2} =\dfrac{\pi^2}{6}