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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. – Did May 9 '11 at 5:40
@Didier: Yeah, i did that but that becomes, more tough. – user9413 May 9 '11 at 5:41
@Didier: Also, i did the standard way of converting definite integral to summation, but that too didn't seem to work. – user9413 May 9 '11 at 5:41
Odd. Power expansion works here. Chandru1: What power series do you get? – Did May 9 '11 at 5:49
Did you do it? Can you show what you did, as precisely as possible (and definitely much less vaguely than in your addendum)? – Did May 9 '11 at 11:26
up vote 3 down vote accepted

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. – Américo Tavares May 11 '11 at 11:57
@Américo The two expansions are the same except for the first term since all derivatives are the same, so there's not much difference. – Adrián Barquero May 11 '11 at 14:52
@Adrián, "all derivatives are the same", that's right. – Américo Tavares May 11 '11 at 21:47

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