Looks Not So Big

a=1+122+132++120162 \large a = 1 + \dfrac1{2^2} + \dfrac1{3^2} + \cdots + \dfrac1{2016^2}

Find the value of a\lfloor a \rfloor .

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

Note by Ayush G Rai
3 years, 4 months ago

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1 vote

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The answer is 1.

Note that if sn=r=1n1r2nN{1}s1=1s_n=\sum_{r=1}^n \frac{1}{r^2} \quad \forall n \in \mathbb{N}-\{1\} \\ s_1=1 then {sn}n=1\{s_n\}_{n=1}^{\infty} is a strictly increasing sequence that which is always less than 2, that is, n<m    sn<smn,mNsn<2nNn<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 sms_m contains more number of positive terms than sns_n.

To prove the next assertion, note that 0<r(r1)<r2rN{1}    1r2<1r(r1)rN{1}    sn<1+r=2n1r(r1)=1+r=2n1r11r=21n    sn<2nN\begin{aligned}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{aligned}

Now, 1=s1<s2016<2    s2016=11=s_1<s_{2016}<2\\\implies \large \boxed{\lfloor s_{2016} \rfloor=1}


Note:

s2016<limnsn=π26<1.645s_{2016}<\lim_{n \to \infty} s_n = \frac{\pi^2}{6} < 1.645

Refer here for the proof of this.

Deeparaj Bhat - 3 years, 4 months ago

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Nice solution..+1.I guessed it anyway

Ayush G Rai - 3 years, 4 months ago

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

Ayush G Rai - 3 years, 4 months ago

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Will try after jee advanced :)

Deeparaj Bhat - 3 years, 4 months ago

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@Deeparaj Bhat How much did you get in jee-main?

Ayush G Rai - 3 years, 4 months ago

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@Ayush G Rai Not so good... 250.

Deeparaj Bhat - 3 years, 4 months ago

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

Ayush G Rai - 3 years, 4 months ago

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I think the answer is 2.

Ayush G Rai - 3 years, 4 months ago

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This is again for NMTC lvl 2 2016

Ankit Kumar Jain - 3 years, 4 months ago

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no it is of 2015

Ayush G Rai - 3 years, 4 months ago

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

Ankit Kumar Jain - 3 years, 4 months ago

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@Ankit Kumar Jain even I appeared

Ayush G Rai - 3 years, 4 months ago

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@Ayush G Rai How many could you solve??

Ankit Kumar Jain - 3 years, 4 months ago

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@Ankit Kumar Jain same as you...1 and a half

Ayush G Rai - 3 years, 4 months ago

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@Ankit Kumar Jain how much problems could you solve??

Ayush G Rai - 3 years, 4 months ago

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

Ankit Kumar Jain - 3 years, 4 months ago

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