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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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    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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    @Adriá$n$, "all derivatives are the same", that's right.2011-05-11