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

Find the value of \(\lfloor a \rfloor \).

**Notation**: \( \lfloor \cdot \rfloor \) denotes the floor function.

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## Comments

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TopNewestThe answer is 1.

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, \[n<m \implies s_n<s_m \quad n, m \in \mathbb{N} \\ s_n<2 \quad \forall n \in \mathbb{N}\]

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 \[\begin{align}0&<r(r-1)<r^2 \quad \forall r \in \mathbb{N}-\{1\} \\\implies \frac{1}{r^2} &< \frac{1}{r(r-1)} \quad \forall r \in \mathbb{N}-\{1\} \\\implies s_n&<1+\sum_{r=2}^n \frac{1}{r(r-1)}\\&=1+\sum_{r=2}^n \frac{1}{r-1}-\frac{1}{r}\\&=2-\frac{1}{n}\\\implies s_n&<2 \quad \forall n \in \mathbb{N} \end{align}\]

Now, \[1=s_1<s_{2016}<2\\\implies \large \boxed{\lfloor s_{2016} \rfloor=1}\]

Note:\[s_{2016}<\lim_{n \to \infty} s_n = \frac{\pi^2}{6} < 1.645\]

Refer here for the proof of this.

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

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Will try after jee advanced :)

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Nice solution..+1.I guessed it anyway

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This is again for NMTC lvl 2 2016

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no it is of 2015

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

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I think the answer is 2.

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