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Can anyone help me in changing the integral into the given form: $$\lim_{n \to \infty}n^{2} \Biggl(\ \ \int\limits_{0}^{1} \sqrt[n]{1+x^{n}} \ \text{dx}-1 \Biggr) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}$$

Once this is done, we know that the integral converges to $\frac{\pi^2}{12}$.

Added: One can generally see that $$(1+x)^{a} = \sum\limits_{k=0}^{\infty} { a \choose k} x^{k}; \qquad x \in [0,1], \ a \in (0,1)$$

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    Could try to expand $\sqrt[n]{1+x^n}$ as a power series and see what happens.2011-05-09
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    @Didier: Yeah, i did that but that becomes, more tough.2011-05-09
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    @Didier: Also, i did the standard way of converting definite integral to summation, but that too didn't seem to work.2011-05-09
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    I deleted my suggestion, since you seem to indicate in your reply to Didier that it was unsuccessful...2011-05-09
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    @J.M: No problem. Suggestions are always useful.2011-05-09
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    Odd. Power expansion works here. Chandru1: What power series do you get?2011-05-09
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    You mean power series expansion along with term by term integration2011-05-09
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    Did you do it? Can you show what you did, as precisely as possible (and definitely much less vaguely than in your addendum)?2011-05-09
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    Chandru: Yes, you can start with that and additionally expand the binomial coefficients to gamma functions.2011-05-09
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    @J.M: Not clear sorry. How does Gamma function come into play.2011-05-09
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    @J.M: Moreover, bringing the series $\frac{1}{n^2}$ seems to be a big issue.2011-05-09
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    Your binomial coefficients can be turned to a ratio of factorials/gamma functions...2011-05-09

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Express $(1+x)^{1/n}-1$ as power series (over $k$) by using $\binom{a}{k}=\frac{a(a-1)\dots (a-k+1)}{k!}$ for $a=1/n$. Integrate termwise.

Now think about why you can interchange limit and summation and take the limit for each summand.

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    It seems to me more natural to expand $(1+x)^{1/n}$ rather than $(1+x)^{1/n}-1$. Most likely I do not fully understand your answer.2011-05-11
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    @Américo The two expansions are the same except for the first term since all derivatives are the same, so there's not much difference.2011-05-11
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    @Adrián, "all derivatives are the same", that's right.2011-05-11