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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 Rai
1 year, 2 months ago

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The 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. · 1 year, 2 months ago

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Try to answer other problems of the set.You are tooooo good in solving problems. · 1 year, 2 months ago

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Will try after jee advanced :) · 1 year, 2 months ago

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How much did you get in jee-main? · 1 year, 2 months ago

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Not so good... 250. · 1 year, 2 months ago

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That is pretty good. well i have a long way to go.I'm still in 9th std going to the 10th. · 1 year, 2 months ago

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Nice solution..+1.I guessed it anyway · 1 year, 2 months ago

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This is again for NMTC lvl 2 2016 · 1 year, 2 months ago

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no it is of 2015 · 1 year, 2 months ago

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Oh sorry I meant the previous year IE 2015....I know because I appeared... · 1 year, 2 months ago

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how much problems could you solve?? · 1 year, 2 months ago

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1 and a half....definitely not so good...One of my friends solved about 3 and qualified · 1 year, 2 months ago

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even I appeared · 1 year, 2 months ago

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How many could you solve?? · 1 year, 2 months ago

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same as you...1 and a half · 1 year, 2 months ago

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I think the answer is 2. · 1 year, 2 months ago

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