1
$\begingroup$

I have this statement:

Prove by induction that $\displaystyle \sum_{k = 1}^n\frac{1}{k^2} \leq 2 -\frac{1}{n}$ for $n \in \mathbb{N}$

My attempt was:

Base case: $\frac{1}{1} \leq 1$

Assume that $\displaystyle \sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$ is true for some $m \in \mathbb{N}$

To prove: $\displaystyle \sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$

Since $$\sum_{k = 1}^m\frac{1}{k^2} \leq 2 -\frac{1}{m}$$

$$\sum_{k = 1}^m\frac{1}{k^2} + \frac{1}{(m+1)^2}\leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$

$$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m} + \frac{1}{(m+1)^2}$$

Now, I'll prove that (1) $$\displaystyle 2 -\frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m + 1}$$

Developing this inequation, gives $$m^2 + 2m + 1 \geq m^2 + 2m$$ getting that (1) is true.

And by transitivity $$\sum_{k = 1}^{m+1}\frac{1}{k^2} \leq 2 -\frac{1}{m+1}$$ was proved.

But I don't know if this is a valid induction proof and if not, what is the type of this proof?

Thanks in advance.

$\endgroup$
  • $\begingroup$ Yes it is an inductive proof and it looks correct to me. $\endgroup$ – sudeep5221 Apr 16 at 1:53
  • $\begingroup$ is not strictly neccessary get the some side(RHS or LHS) ? in this case, i develop a inequality seperate and after incorporate that proof to the original proof. Is this still valid for formal induction proof? $\endgroup$ – Eduardo Sebastián C. Apr 16 at 1:56
  • $\begingroup$ It is just the way you have written the proof. It could have equivalently been written as $\displaystyle 2 - \frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} + \frac{1}{m+1} - \frac{1}{m} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} - \frac{1}{m(m+1)} + \frac{1}{(m+1)^2} \leq 2 - \frac{1}{m+1} + \frac{1}{(m+1)}\left(\frac{1}{m+1} - \frac{1}{m} \right) \leq 2 - \frac{1}{m+1} $ $\endgroup$ – sudeep5221 Apr 16 at 2:00
1
$\begingroup$

Let me show you a different (more direct) way to prove this inequality. Note that $k(k-1)<k^2$ and therefore $$\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}$$ for all $k>1.$
Now for $n>1$ we have $$\sum_{k=1}^n\dfrac{1}{k^2}\le1+\sum_{k=2}^n\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)=1+\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\cdots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right).$$ Simplify the RHS and see it for your self :)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ For more closer approximation of the series, use the inequality $k^2-1<k^2$ and repeat this same procedure with the partial fraction decomposition $$\dfrac{1}{k^2-1}=\dfrac1{2(k-1)}-\dfrac1{2(k+1)}.$$ $\endgroup$ – Bumblebee Apr 16 at 3:33
0
$\begingroup$

If we know that it is valid for $n$, we only need to show that $\frac{1}{(n+1)^{2}}<\frac{1}{n}-\frac{1}{n+1}$

Therefore

$$ \sum_{k=1}^{n}{\frac{1}{k^{2}}}+\frac{1}{(n+1)^{2}}<2-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1} $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.