Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$

Question

While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.

I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms $n = 1, n=2$ I get:

for $n = 1$, $\frac{1}{1^2} = 1 < 2$ for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$

Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$ But now what? If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$. I am stuck.

Thanks for your thoughts,

Brian

Answer 1

Daniel's comment gives you the answer (the easiest way, imo):

$$\frac1{n^2}\le \frac1{n(n-1)}=\frac1{n-1}-\frac1n\implies$$

$$\sum_{k=2}^n\frac1{k^2}\le\sum_{k=2}^n\left(\frac1{n-1}-\frac1n\right)=1-\frac12+\frac12-\frac13+\ldots+\frac1{n-1}-\frac1n=1-\frac1n\xrightarrow[n\to\infty]{}1$$

Now only add the first summand in your series to the above and we're done.

Answer 2

I'm inclined to agree that the telescoping series is the slickest way to get the inequality, but here's one more approach:

$$\begin{align} \sum_{n=1}^\infty {1\over n^2}&=1+{1\over4}+{1\over9}+{1\over16}+{1\over25}+{1\over36}+{1\over49}+{1\over64}+\cdots\cr &\lt 1+{1\over4}+{1\over4}+{1\over16}+{1\over16}+{1\over16}+{1\over16}+{1\over64}+\cdots\cr &=1+{1\over2}+{1\over4}+{1\over8}+\cdots\cr &=2\cr \end{align}$$

Answer 3

One way is to look at the lower Riemann sum with width $1$ under the graph of $f(x) = {1 \over x^2}$, from $x = 1$ to $x = \infty$. Since ${1 \over x^2}$ is decreasing, the lower Riemann sum will be $\sum_{n=2}^{\infty} f(n) = \sum_{n=2}^{\infty} {1 \over n^2}$, which must be less than the integral $\int_1^{\infty} {1 \over x^2}\,dx = 1$. Thus adding $1$ to this inequality gives the result you seek.

Answer 4

Let $S_n(p) = \displaystyle \sum_{k=1}^n \dfrac1{k^p}$ and $S(p) = \lim_{n \to \infty} S_n(p)$. We then have $$S_{2n+1}(p) = \sum_{k=1}^{2n+1} \dfrac1{k^p} = 1 + \sum_{k=1}^n \left(\dfrac1{(2k)^p} + \dfrac1{(2k+1)^p}\right) < 1 + \sum_{k=1}^n \dfrac2{(2k)^p} = 1 + \dfrac{S_n(p)}{2^{p-1}}$$ Letting $n \to \infty$, we get that $$S(p) \leq 1 + \dfrac{S(p)}{2^{p-1}} \implies S(p) \leq \dfrac{2^{p-1}}{2^{p-1}-1}$$

Answer 5

Hint Prove instead that

$$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2-\frac{1}{n}$$

Interesting, this stronger exercise is an easy induction problem, while the weaker result you mentioned cannot be proven directly by induction for obvious reasons (LHS increases, RHS is constant < --- in the stronger version both sides are increasing, this issue is fixed).

Answer 6

$$\sum_{n=1}^\infty{1\over n^2}<1+\sum_{n=2}^\infty{1\over n^2-{1\over4}}= 1+\sum_{n=2}^\infty\left({1\over n-{1\over2}}-{1\over n+{1\over2}}\right)=1+{2\over3}={5\over3}\ .$$ By the way: Since the left hand side is $={\pi^2\over6}$ you obtain from this the estimate $\pi^2<10$, which is not bad. In fact $\pi^2\doteq9.87$.