Proof of inequality

[Promilla]

Could you please help me with this inequality. I cannot prove it. I was trying to do this for n but it turned out that for n it is not true.
How should I do this? Where should I start?

\(\displaystyle \sum_{n=1}^{10000}1/n^{2}<2-1/10001\)

 

[pka]

Could you please help me with this inequality. I cannot prove it. I was trying to do this for n but it turned out that for n it is not true.
\(\displaystyle \sum_{n=1}^{10000}1/n^{2}<2-1/10001\)

I have no idea what whoever assigned the question expects you to do,

However, \(\displaystyle \displaystyle\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}\)

You sum is just one of those increasing approximating sums and clearly \(\displaystyle \dfrac{{{\pi ^2}}}{6} < 2 - \frac{1}{{10001}}\)

 

[Promilla]

Now show that \(\displaystyle 2 - \dfrac{1}{k+1} + \dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{(k+1) + 1}} = 2 - \dfrac{1}{k+2}.\)

But this is not true.

 

[Deleted member 4993]

But this is not true.

It is not???

Can you find the value of k>1 for which that is not true?

 

[Promilla]

for example 2 .

 

[pka]

The proof is not obvious, at least not to me. I erred. I have worked much of the afternoon and have not found a proof.

May I suggest that you review reply #2.
This particular series is such a well known result.
The partial sums are increasing to the final value \(\displaystyle \frac{\pi^2}{6}\).

 

[pka]

That does not represent a proof.

Why do you say that?
Just as I don't, you have absolutely no idea what method the author of the question wanted?
Without any more instructions, that is absolutely a proof.

 

[Deleted member 4993]

for example 2 .

Now show that

You are correct Promila - I had the inequality sign reversed. Thus:

\(\displaystyle 2 - \dfrac{1}{k+1} + \dfrac{1}{(k+1)^2} \ = 2 - \dfrac{k}{(k+1)^2} \ = \ 2 - \dfrac{1}{k+2+\dfrac{1}{k}} > \ 2 - \dfrac{1}{k+2}\)

.

 

[Promilla]

Thank you all. I think it will be sufficient. Thanks.