Find the limit of $g_n=1+{1\over 2^2}+\cdots+{1\over n^2}$ 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 ?
 A: 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.
A: 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.
A: 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. 
A: 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.
A: 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.
