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Suppose i need to prove that $\frac{1}{2^2}+\frac{1}{3^2}...+\frac{1}{n^2}<1-\frac{1}{n}$

So in the step of $n+1$, the right side becomes $<1-\frac{1}{n+1}$ or is it: $<1-\frac{1}{n}-\frac{1}{n+1}$? i guess it's the first one but why? i mean in the left side it's in addition to the first term $(\frac{1}{n^2})$ which is $\frac{1}{(n+1)^2}$.

*This question is mostly because i saw answers for this where they add $\frac{1}{(k+1)^2}$ to both sides, which is complex than just to write : $1-\frac{1}{n}=\frac{n-1}{n}$ and in the step of $n+1$ it just becomes: $\frac{n}{n+1}$ from there it's simple so i didn't understand why to add that term unless i'm missing something...

Thank you!

Thanks you.

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For the induction step, you assume the following: $$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} < 1-\frac 1 n$$ and we want to prove the following: $$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}+\frac{1}{(n+1)^2} < 1-\frac 1 {n+1}$$ Notice how we simply substituted $n+1$ into the original expression. We didn't do anything else; simply substituted.

Now, what the people in the answer did is that they started with the first expression and then added both sides by $\frac{1}{(n+1)^2}$ in order to get closer to the second inequality by changing the left-hand side. They started with the first inequality and transformed it into the second in order to prove the second inequality.

Remember, to prove something, we being with our assumptions and then use logical and mathematical manipulations to get a statement equivalent to our conclusion. By adding $\frac{1}{(n+1)^2}$ to both sides of the assumption inequality, they got an inequality that looked more like the conclusion and thus got closer to proving it. They then used simplifications and Transitive Property of Inequality to transform the right side into what they needed, but their first step was to transform the left side of the assumption inequality into the left side of the conclusion inequality.

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  • $\begingroup$ that i understand (the logic), but why to do it if it's direct by what i suggested here? and secondly if you just put $n+1$ as you said so why on the left side it not only $\frac{1}{(n+1)^2}$? i understand that you have to term before it which is $\frac{1}{n^2}$ but in the right side we didn't consider that and just put $n+1$, that i don't understand. $\endgroup$ – bony Jun 19 '16 at 21:32
  • $\begingroup$ @bony The above is not a proof. All I was showing you is the assumption inequality and then the conclusion inequality below it. In a proof, the assumption inequality comes first, then you add $\frac{1}{(n+1)^2}$ to both sides so the inequality becomes: $$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} < 1-\frac 1 n+\frac{1}{(n+1)^2}$$ Then, the different answers used simplifications to simplify that to show that: $$1-\frac 1 n+\frac{1}{(n+1)^2} < 1-\frac{1}{n+1}$$ Then, finally, they used Transitive Property of Inequality to get the conclusion inequality at the end. $\endgroup$ – Noble Mushtak Jun 19 '16 at 21:35
  • $\begingroup$ yes that i understand. just didn't understand why not just to do it direct it's shorter and simpler. so i thought maybe i'm getting it all wrong, and im still not getting the step without adding that term. why by putting $n+1$ you don't do nothing but substituted in the right side but on the left side you keep the last term and and new term . $\endgroup$ – bony Jun 19 '16 at 21:40
  • $\begingroup$ @bony Oh! This makes your confusion a lot easier to understand. Rigorously, the left side is actually: $$\sum_{i=2}^n \frac{1}{i^2}$$ Now, for the statement for $n+1$, we just substitute $n+1$ for $n$, so it becomes: $$\sum_{i=2}^{n+1} \frac{1}{i^2}$$ Now, as you see, this is the same as just adding $\frac{1}{(n+1)^2}$ to the left side because we just added a new term to the series. However, this does not apply to the right side because the right side is not a series. $\endgroup$ – Noble Mushtak Jun 19 '16 at 21:44
  • $\begingroup$ i got it, so the right side is just a term (terms) not a series. thanks! $\endgroup$ – bony Jun 19 '16 at 22:44
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Your induction should be as follows: assume that this property holds for all $k≤n$. Then your inductive step for $n+1$ is: $$\frac{1}{3^2} +\ldots +\frac{1}{n^2}+\frac{1}{(n+1)^2}<\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}$$ This holds as you can verify for yourself that $\frac{1}{2^2}>\frac{1}{(n+1)^2}$ for any $n≥2$. Since $\frac{1}{n}>\frac{1}{n+1}$, and by our assumption, we now have that the equation is: $$\frac{1}{3^2} +\ldots +\frac{1}{(n+1)^2}< \frac{1}{2^2}+\ldots +\frac{1}{n^2}<1-\frac{1}{n}<1-\frac{1}{n+1}$$ And our inductive step is shown.

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While this does not answer the specific question in the OP, I thought it would be instructive to present an approach that develops the inequality directly. To that end, we proceed.

Note that we have

$$\begin{align} \frac{1}{k^2}& \le \frac{1}{k(k-1)}\\\\ &=\frac{1}{k-1}-\frac{1}{k} \end{align}$$

Therefore, we can write

$$\begin{align} \sum_{k=2}^n\frac{1}{k^2}& \le \sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right) \tag 1\\\\ &=1-\frac{1}{n} \tag 2 \end{align}$$

where we used the telescoping nature of the summand to go from $(1)$ to $(2)$. And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Aug 12 '16 at 1:47

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