Can someone tell me if I am doing this right and what is the next step (mathematical induction of a sum?)?

Question

I am asked to show that$$\sum_{i=1}^n \frac{1}{i^2} \le 2-\frac{1}{n}.$$

Using my notes from a discrete math class, I figure I should use mathematical induction(?) and this is what I have so far:

Base Case: $n=1$

Suppose $$\sum_{i=1}^k \frac{1}{i^2} \le 2 -\frac{1}{k}$$ Then $$\sum_{i=1}^{k+1} \frac{1}{i^2} = \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} $$ $$=2 - \frac{1}{k}+ \frac{1}{(k+1)^2}$$

This is where I get lost. I'm not sure if this last part is correct or what to do next. (I'm not sure if I am even doing this correctly)

You made a few mistakes in the last two equations, the correct version of the induction step is $$\sum_{i=1}^{k+1} \frac{1}{i^2} = \sum_{i=1}^k \frac{1}{i^2} + \frac{1}{(k+1)^2} \leq 2-\frac{1}{k} + \frac{1}{(k+1)^2}$$ Can you take it from here? — Andreas Tsevas, Sep 06 '22 at 21:41
Does this answer your question? Show that $\sum_{i=1}^n \frac{1}{i^2} \le 2 - \frac{1}{n}$ - found using an Approach0 search. Other duplicates include Prove that for $ \displaystyle n≥1, \sum_ {i=1} ^{ n} \dfrac 1 {i^2} ≤ 2 - \dfrac 1 n$, Standard Problem on Mathematical-Induction, ... — John Omielan, Sep 06 '22 at 22:15
@Andreas ah I see where I messed up in my notes now thank you. But I'm still not sure where to go next. Something to do with the denominators for the right hand side? — Ether, Sep 06 '22 at 22:20
@Ether (cont.) Proof by induction involving inequalities and summation, and Is $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ correctly proved by induction?. Also, your problem is the first one in How does $\frac{-1}{k}+\frac{1}{(k+1)^2} \leq \frac{-1}{k+1}$ helps to show that $\sum\limits_{i=1}^k\frac1{i^2}\leq \ 2-\frac{1}{k}$, and a solution to it is asked about in Prove summation inequality with induction. This is from just the first page of results, so there's likely more. Last, but not least, welcome to Math SE. — John Omielan, Sep 06 '22 at 22:22
@JohnOmielan Actually yeah thank you! I messed up by having the common denominator as k(k+1)^2 instead of just (k+1)^2. Thank you for the links my searches weren't turning up anything but I my not have been phrasing it right. — Ether, Sep 06 '22 at 22:30
@Ether You're welcome. Note that site searches, and even using general search engines like Google or Bing, often don't work very well with searching math equations, mainly due to things like somewhat different symbols, ordering, etc., being used. Instead, as I mentioned, I used Approach0, which handles MathJax text, including equations & inequalities, quite well. For more information about using that search engine, as well as searching on this site in general, please read How to search on this site?. — John Omielan, Sep 06 '22 at 22:35

Can someone tell me if I am doing this right and what is the next step (mathematical induction of a sum?)?

0 Answers0