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How do I prove that $\sum_{n=1}^{\infty}\frac{2^{n}}{1+2^{n}}$ converges or not using comparison test?

I tried the ratio test, and I couldn't use it because $\frac{a_{n+1}}{a_{n}}\rightarrow 1$.

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  • $\begingroup$ It diverges by the divergence test $\endgroup$ – user223391 Mar 4 '17 at 23:42
  • $\begingroup$ Cant you use the limit comparison test $\endgroup$ – TheGathron Mar 4 '17 at 23:57
  • $\begingroup$ Why would you do that? $\endgroup$ – user223391 Mar 5 '17 at 0:00
  • $\begingroup$ It's an exercise I couldnt solve $\endgroup$ – TheGathron Mar 5 '17 at 0:01
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    $\begingroup$ I (and carmichael) told you how to solve it. $\endgroup$ – user223391 Mar 5 '17 at 0:02
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Recall that if $\sum_{n}a_n$ is a convergent series then $\lim_{n\to\infty}a_n=0$.

In this case, since $\lim_{n\to\infty}\frac{2^n}{2^n+1}=1\neq 0$, the series diverges.

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Since $2^n \ge 1$, $1+2^n \le 2\cdot 2^n$, so $\frac{2^{n}}{1+2^{n}} \ge \frac12$.

Therefore $\sum_{n=1}^{m}\frac{2^{n}}{1+2^{n}} \ge \sum_{n=1}^{m} \frac12 =\frac{m}{2} $, so the sum diverges as $m \to \infty$.

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    $\begingroup$ For a constructive approach (+1). Judging by the votes, applying the divergence test was viewed as quite remarkable. $\endgroup$ – RRL Mar 5 '17 at 2:16
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    $\begingroup$ I think that it being submitted an hour earlier made the difference. $\endgroup$ – marty cohen Mar 5 '17 at 2:31

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