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Show that the given sequence $\langle g_n\rangle$ is monotonically increasing, is bounded and $\lim g_n\le2$. Given that

$$g_n=1+{1\over 2^2}+\cdots+{1\over n^2}$$

Showing that the given sequence is monotonically increasing is quite straight forward. How do i show that it is bounded and find its limit ?

Michael Hardy
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Aman Mittal
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  • It is a general fact that if a sequence of real numbers is monotonically increasing and bounded above, then it converges to its least upper bound. However, determining what that least upper bound is may be difficult in particular cases. In this case, to explicitly determine the least upper bound requires some sophisticated Mathematics however you do it ( at least to my knowledge). – Geoff Robinson May 12 '14 at 17:55
  • can this be converted into a GP with an inequality ? – Aman Mittal May 12 '14 at 17:59
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    By looking at areas you can show that $\frac{1}{2^2}+\cdots+\frac{1}{n^2}\lt \int_1^n \frac{1}{x^2},dx\lt 1$. – André Nicolas May 12 '14 at 18:01
  • @AmanMittal: Yes, you can do a comparison with a suitable GP. – André Nicolas May 12 '14 at 18:03
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    To find the limit: http://math.stackexchange.com/questions/8337/different-methods-to-compute-sum-limits-n-1-infty-frac1n2 – Hans Lundmark May 12 '14 at 18:05
  • @GeoffRobinson: I wrote out a solution. Same idea works for $\sum \frac{1}{n^p}$, $p\gt 1$. – André Nicolas May 12 '14 at 18:17

5 Answers5

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To show that it is bounded, replace each summand $\frac1{k^2}$ ($k>1$) with the larger summand $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$, which results in a telescope sum. You won't find the limit so easily - even if I tell you that it is $\frac{\pi^2}6$, this will not give you the slightest idea how to arrive at that result.

Pedro
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Hagen von Eitzen
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    That's less than 1. Did you mean $\pi^2/6$? – Christopher Liu May 12 '14 at 17:59
  • @ChristopherLiu The first term is omitted from the telescoping sum, so the bound obtained is actually $2$, which is the requested bound. This is almost certainly the proof they wanted. – user2566092 May 12 '14 at 18:01
  • I meant the part where he says the limit is $6/\pi^2$. How can this be? – Christopher Liu May 12 '14 at 18:05
  • ok, so going this way i find that the sequence is bounded above by $2-{1\over n}$ so using the fact that limit of a monotonically increasing sequence is equal to the least upper bound, $g_n \le 2$ does that look good ? – Aman Mittal May 12 '14 at 18:08
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    @ChristopherLiu You're right, that part is a typo. The classical result is that the limit is $\pi^2/6$, like you said. But the argument for the bound of $2$ is totally legit. – user2566092 May 12 '14 at 18:09
  • @AmanMittal Yes exactly, all partial sums are bounded by $2 - 1/n$ so you get $2$ as the bound on the sequence, exactly as requested by the problem. This is almost certainly the solution they had in mind since they said to prove $2$ is a bound. – user2566092 May 12 '14 at 18:11
  • @user2566092 Thanks !! – Aman Mittal May 12 '14 at 18:11
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Consider $$S_n(p) = \sum_{k=1}^n \dfrac1{k^p}$$ It is easy to show that $S_n(p)$ is monotone increasing in $n$. We then have \begin{align} S_{2n+1}(p) & = \sum_{k=1}^{2n+1} \dfrac1{k^p} = 1 + \sum_{k=1}^n\left(\dfrac1{(2k)^p} + \dfrac1{(2k+1)^p}\right)\\ & \leq 1 + \sum_{k=1}^n\left(\dfrac1{(2k)^p} + \dfrac1{(2k)^p}\right)= 1+\sum_{k=1}^n\dfrac{2^{1-p}}{k^p} = 1+2^{1-p} S_n(p) = 1+2^{1-p} S_{2n+1}(p) \end{align} This gives us $$S_{2n+1}(p) \leq \dfrac1{1-2^{1-p}}$$ Taking $p=2$, gives the result you are after.

Srinivasa Ramanujan
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OP has asked about comparison with a geometric progression. Yes, it can be done, though of course not term by term.

The two terms $\frac{1}{2^2}$ and $\frac{1}{3^2}$ have sum $\lt 2\cdot \frac{1}{2^2}=\frac{1}{2}$.

The four terms $\frac{1}{4^2}$ to $\frac{1}{7^2}$ have sum $\lt 4 \cdot \frac{1}{4^2}=\frac{1}{4}$.

The eight terms from $\frac{1}{8^2}$ to $\frac{1}{15^2}$ have sum $\lt \frac{1}{8}$.

And so on.

Remark: The same idea leads to the Cauchy Condensation Test.

André Nicolas
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You could use the fact that $$\zeta(2) = \sum_{n}\frac{1}{n^2} = \frac{\pi^2}{6}$$ There are many proofs of this. See http://en.wikipedia.org/wiki/Basel_problem.

Mayank Pandey
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As pointed out in another answer, you can get an upper bound by replacing $1/k^2$ with $1/k(k-1)$ for $k > 1$ and get a telescoping sum. This must be the solution that they had in mind, because the value of the telescoping sum is equal to $2$ after you add the first term, which is the requested bound in the problem.

user2566092
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