# Determine if $\sum_{n=1}^{\infty}\frac{2^{n}}{1+2^{n}}$ converges

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 $\frac{a_{n+1}}{a_{n}}\rightarrow 1$.

• It diverges by the divergence test – user223391 Mar 4 '17 at 23:42
• Cant you use the limit comparison test – TheGathron Mar 4 '17 at 23:57
• Why would you do that? – user223391 Mar 5 '17 at 0:00
• It's an exercise I couldnt solve – TheGathron Mar 5 '17 at 0:01
• I (and carmichael) told you how to solve it. – user223391 Mar 5 '17 at 0:02

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.
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 $m \to \infty$.