Help with proof using induction: $1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ [duplicate]

I am having trouble with the following proof:

For every positive integer $n$: $$1 + \frac{1}{4} + \frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$$

My work: I have tried to add $\frac{1}{(k+1)^2}$ to $2-\frac{1}{k}$ in the inductive step and reduce it down to $2-\frac{1}{k+1}$ but cannot do so. I am beginning to think that my entire approach is wrong.

marked as duplicate by Martin Sleziak, jameselmore, quid♦, drhab, Daniel FischerOct 27 '15 at 9:29

• $$n\lt n+1\iff n(n+1)\lt (n+1)^2\iff \dfrac{1}{(n+1)^2}\lt\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$ – Bumblebee Mar 25 '15 at 5:35
• – Martin Sleziak Oct 26 '15 at 14:50

Inductive Hypothesis: suppose $\sum_{k=1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$.

Inductive Step: then suppose $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq 2 - \frac{1}{n} + \frac{1}{(1+n)^2}.$

So it suffices to show that $- \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$.

But this, by simple algebra, is equivalent to $n \leq n+1$, which is obviously true $\forall n \in \mathbb{N}$.

EDIT Proof that $- \frac{1}{n} + \frac{1}{(1+n)^2} \leq - \frac{1}{n+1}$ is equivalent to $n \leq n+1$.

Add $\frac{1}{n}$ to both sides to se that what we need to show is $\frac{1}{(1+n)^2} \leq \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)}.$ Multiply both sides by $(n+1)$. Then take reciprocals and remember to swap the inequality sign.

• Thank you. It makes much more sense now. – mrQWERTY Mar 19 '14 at 21:13
• Good, you're welcome :) You can use the 'tick' button to accept my answer in that case, if you wish... ;) – Frank Mar 19 '14 at 21:14
• Hi, I have a problem trying to arrive at $n \leq n+1$. When trying to reduce the expression, I always arrive at $n \leq \frac{(n+1)^2}{n+2}$. I do not know how to reduce that down further. Please pardon me for my newbness. – mrQWERTY Mar 19 '14 at 21:47
• See the edit to my post above. – Frank Mar 19 '14 at 21:51
• Awesome, thank you. – mrQWERTY Mar 19 '14 at 21:54

Hint: $\frac{1}{n^2} < \frac{1}{(n-1)} - \frac1n$, and telescope.

In the induction step observe that $1$+$1/4$ +...+$1/k^{2}$ + $1/(k+1)^{2}$$\leq 2- 1/k + 1/(k+1)^{2} Now 1/(k+1)^{2} \leq 1/k(k+1)=1/k-1/(k+1). Use this inequality in previous to obtain the answer. Others have commented on using induction. Here's another approach just 'cause: We know the infinite sum is equal to \frac{\pi^2}{6} which is approximately 1.644, so we know it is true for n > 2 (infinite sum > partial sum with all positive terms) So we can just hand check n = 1, 2: Indeed, 1 \leq 1 and \frac{5}{4} \leq \frac{3}{2} • talk about sledgehammers and molehills! – Frank Mar 19 '14 at 21:28 • it was never said that you couldn't use that and the two questions aren't really that related so figured why not! – MCT Mar 19 '14 at 21:30 • you could probably use Fermat's theorem to prove lots of elementary things ;) – Frank Mar 19 '14 at 21:31 • haha, yeah. There was a recent homework post to prove that (a+1)^3 \neq a^3 + (a-1)^3 for any positive integer a > 2, and someone jokingly commented with a wikipedia article to FLT. – MCT Mar 19 '14 at 21:33 • hahahahah excellent :) – Frank Mar 19 '14 at 21:34 Just thought I should comment on the inequality$$-\frac{1}{n} + \frac{1}{(n+1)^2} \leq -\frac{1}{n+1}$$that needs to be shown in the induction argument. If we let f(x) = 1/x then this is equivalent to showing f(n+1)-f(n) \leq f'(n+1). But$$f(n+1)-f(n) = \int_n^{n+1} f'(x) dx \leq f'(n+1)$$since$f'(x) = -1/x^2$is increasing for positive$x$. • It doesn't need to be shown in any sophisticated way at all. As I show, it is immediately equivalent to the inequality$n \leq n+1$. – Frank Mar 19 '14 at 22:59 • Nothing needs to be shown in any way at all! I'm just giving another solution. (Besides, it's only calculus.) From this we could generalize the above problem to proving that$\sum_{k=1}^n f'(k) \leq C + f(n)$for an appropriate constant$C$, or something like that, where$f'(x)\$ is increasing. – abnry Mar 19 '14 at 23:50