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A have a series $\displaystyle \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}$ I would like to know if it is possible to find a large finite integer n such that:

$\displaystyle \sum_{k=n}^{\infty}\frac{1}{(2k-1)^2}<\epsilon$, $\displaystyle \forall \epsilon \in \mathbb{R}^{+}$

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    Please, do not change your question in such a way *after* an answer has been posted. – 2017-01-17
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    It is just the definition of "convergence" for a series, and yes, $$\sum_{k\geq 1}\frac{1}{(2k-1)^2}$$ is convergent, for instance by comparison with a telescopic series. Its exact value is $\frac{\pi^2}{8}$. – 2017-01-17
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    @Jack D'Aurizio look at the quantifiers, the "natural" parsing of the sentence makes them wrong for the definition of convergence you mention. – 2017-01-17
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    @ClementC.: you are right, but I also think the original question had the quantifiers in the most natural order $$\forall\varepsilon>0,\exists N\in\mathbb{N}: \sum_{n\geq N}\frac{1}{(2n-1)^2}<\varepsilon.$$ Otherwise, the question is really trivial. – 2017-01-17
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    The original question had no epsilon, so it's unclear to me what the OP originally intended (that's why my answer covers all 3 variants). – 2017-01-17
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No (for the original question, with equality to zero): $\exists n, \sum_{k=n+1}^\infty \frac{1}{(2k-1)^2} = 0$.

You are summing non-negative terms. Having the sum equal to $0$ would imply every single term is zero. But every term is positive.

With more symbols (but the same argument): assume by contradiction such a $n\geq 1$ exists. Then $$ 0 = \frac{1}{(2n-1^2} + \sum_{k=n+1}^\infty \underbrace{\frac{1}{(2k-1^2}}_{\geq 0} \geq \frac{1}{(2n-1^2} > 0 $$ leading to a contradiction.

No, for the new question with the quantifiers: $\exists n, \forall \varepsilon > 0,\ \sum_{k=n+1}^\infty \frac{1}{(2k-1)^2} < \varepsilon$.

Indeed, this is then the same question as before, as this implies equality to $0$ (just take the limits on both sides as $\varepsilon \to 0$).

Yes, for the different question inverting the quantifiers: $\forall \varepsilon > 0, \exists n,\ \sum_{k=n+1}^\infty \frac{1}{(2k-1)^2} < \varepsilon$.

Your series converges, so the remainder tends to zero by definition. That is, as in addition the terms of the series are all non-negative, $$ 0 \leq \sum_{k=n}^\infty \frac{1}{(2k-1)^2} \xrightarrow[n\to\infty]{} 0 $$ which will give you what you want (by recalling the definition of a limit).

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    Clement C. see my edit – 2017-01-17
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    So, you basically entirely changed the question after the answer? That's not good practice. – 2017-01-17
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    Oh i'm Sorry!!! – 2017-01-17
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    You don't need the contradiction argument. The last two inequalities alone give you the direct proof. – 2017-01-17
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    @IsraelMeirelesChrisostomo I have edited my anser -- see above. – 2017-01-17
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    It is in fact the same question: if a non-negative number is less than $\;\epsilon\;$ , **for all** $\;\epsilon >0\;$ , then it **must** be zero... – 2017-01-17
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    @DonAntonio Yes, see what I wrote. – 2017-01-17
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    Excellent answer to *several* questions... +1 – 2017-01-17