Given following sequence: $$x_{n}=\frac{1}{2+1}+\frac{1}{2^2+1}+...+\frac{1}{2^n+1}$$ I have to prove that this sequence has a limit. So I think first of all I should prove that this is convergent sequence. But I have absolutely no idea how to do this because I haven't seen any examples with such complicated sequence before. So can you give me some hint how to do my first step? Any help will be appreciated.
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1$\begingroup$ Hint $\frac{1}{2^k+1} < \frac{1}{2^k}$ and $\frac{1}{2} + \frac{1}{2^2}+\ldots = 1$ $\endgroup$– WintherSep 28, 2014 at 21:24
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3 Answers
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Hint: Use that $\dfrac{1}{2^n+1}\le \dfrac{1}{2^n}$ to show it's bounded and show it's increasing.
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$\begingroup$ Maybe you should write $\lt$ instead of $\le$? $\endgroup$– k1berSep 28, 2014 at 21:46
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$\begingroup$ @KiberPrestupnik $\le$ is good enough. $\endgroup$– user121880Sep 28, 2014 at 22:14
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Comparison test with the sequence $y_n=\frac{1}{2}+\frac{1}{2^2}+...\frac{1}{2^n}$.
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$\textbf{Hint:} $Prove that $x_n$ is a Cauchy sequence. Then $x_n$ has a limit.
