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.

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

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

\(\text{Therefore, }\lfloor a \rfloor=\boxed{1}\) – Akshat Sharda · 1 year, 2 months ago

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\(\large\ \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } = \frac { \pi ^{ 2 } }{ 6 } }\) ? – Priyanshu Mishra · 1 year, 2 months ago

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this. – Akshat Sharda · 1 year, 2 months ago

SeeLog in to reply

– Priyanshu Mishra · 1 year, 2 months ago

Thanks ,now i understood. However , how many points will you give for this problem?Log in to reply

– Akshat Sharda · 1 year, 2 months ago

75 of 400Log in to reply

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. – Aditya Anand · 1 year, 2 months ago

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[a]=1 – Saran .P.S · 1 year, 2 months ago

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Question 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. – Silver Vice · 1 year, 1 month ago

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– Akshat Sharda · 1 year, 1 month ago

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 \(a\).Log in to reply

By the way, how many questions you solved in the exam ? – Priyanshu Mishra · 1 year, 1 month ago

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– Akshat Sharda · 1 year, 1 month ago

I didn't gave the exam :-(Log in to reply

– Silver Vice · 1 year, 1 month ago

Ok, :) , integral value will be one only.Log in to reply

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... – Saran .P.S · 1 year, 2 months ago

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