# Looks Not So Big

$\large a = 1 + \dfrac1{2^2} + \dfrac1{3^2} + \cdots + \dfrac1{2016^2}$

Find the value of $\lfloor a \rfloor$.

Notation: $\lfloor \cdot \rfloor$ denotes the floor function.

Note by Ayush G Rai
3 years, 9 months ago

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This is again for NMTC lvl 2 2016

- 3 years, 9 months ago

no it is of 2015

- 3 years, 9 months ago

Oh sorry I meant the previous year IE 2015....I know because I appeared...

- 3 years, 9 months ago

how much problems could you solve??

- 3 years, 9 months ago

1 and a half....definitely not so good...One of my friends solved about 3 and qualified

- 3 years, 9 months ago

even I appeared

- 3 years, 9 months ago

How many could you solve??

- 3 years, 9 months ago

same as you...1 and a half

- 3 years, 9 months ago

Note that if $s_n=\sum_{r=1}^n \frac{1}{r^2} \quad \forall n \in \mathbb{N}-\{1\} \\ s_1=1$ then $\{s_n\}_{n=1}^{\infty}$ is a strictly increasing sequence that which is always less than 2, that is, $n

Here are the proofs:

The first statement is true as $s_m$ contains more number of positive terms than $s_n$.

To prove the next assertion, note that \begin{aligned}0&

Now, $1=s_1

Note:

$s_{2016}<\lim_{n \to \infty} s_n = \frac{\pi^2}{6} < 1.645$

Refer here for the proof of this.

- 3 years, 9 months ago

Try to answer other problems of the set.You are tooooo good in solving problems.

- 3 years, 9 months ago

Will try after jee advanced :)

- 3 years, 9 months ago

How much did you get in jee-main?

- 3 years, 9 months ago

Not so good... 250.

- 3 years, 9 months ago

That is pretty good. well i have a long way to go.I'm still in 9th std going to the 10th.

- 3 years, 9 months ago

Nice solution..+1.I guessed it anyway

- 3 years, 9 months ago

I think the answer is 2.

- 3 years, 9 months ago