Is every sequence that converges to 0 also has a convergent subsequence?

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Homework Statement


If every subsequence of the sequence a_n converges to 0, prove that a_n itself converges to 0. Is the converse true? Prove it or give a counter example.


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The Attempt at a Solution


I can prove a sequence is convergent if it has a convergent subsequence. What about every sequence is convergent?
 
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This is a similar reasoning to the proof that if xn converges to x, then all subsequences that converge do so to x. You can do this by considering subsequences xj and xk where j is odd and k is even.
 
Look at the definition of convergence. The one I learned was that a_n converges to a number L if and only if given e > 0 there exists an N such that |a_n - L| < e for all n>N. Hint: use the triangle inequality:

a_n - L = a_n + \underbrace{(- b_j + b_j - b_k + b_k)}_{=0} - L<br /> = (a_n - b_j) + (b_j - b_k) + (b_k - L)

which implies

|a_n-L| \le |a_n - b_j| + |b_j - b_k| + |b_k - L|

If you consider the b_j and b_k as subsequences, you should be able to bound the right hand side using the definition of convergence.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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