Hello friends, help me in finding the value of

\(\large\ a = 1 + \frac { 1 }{ { 2 }^{ 2 } } + \frac { 1 }{ { 3 }^{ 2 } } + ... + \frac { 1 }{ { 2015 }^{ 2 } }\) .

Do post solution.

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## Comments

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TopNewestQuestion has asked for value of a, not floor function of a. So a will be equal to approx. 1.64 . Do note that the series is going smaller and smaller.

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In the question he didn't wrote about the integral value but in the original paper it was asked to find the integer part of \(a\).

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Yes, I forgot to write that.

By the way, how many questions you solved in the exam ?

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Ok, :) , integral value will be one only.

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Well,it is obvious if one knows the solution of the basel problem .However this is a much simpler question.Notice that the first term is 1,the next two terms are less than 2×1÷2^2 that is 1/2,the next four terms less than 1/4 and so on.Even if we did this til infinity,the expression would be less than 1+1/2+1/4.... which is 2.Hence the required value is 1.

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Solution :\(\displaystyle \sum^{\infty}_{n=1}\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6} \sim 1.64 \text{ and }\displaystyle \sum^{2015}_{n=1}\dfrac{1}{n^{2}}=a\)

\(\text{Simply by observation, we can conclude }\Rightarrow 1<a<1.64\)

\(\text{Therefore, }\lfloor a \rfloor=\boxed{1}\)

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Thank for a solution. But please explain me how you got this:

\(\large\ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } = \frac { \pi ^{ 2 } }{ 6 } }\) ?

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See this.

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that is now we have 1+....so on .consider 1 the next terms are decreasing values in decimals .so anyway if we add also we won't get more than 1 so the integral part of that expression i.e the[a] is 1...

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[a]=1

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