I want to show that $\sum_{n=1}^\infty {\frac{a^n}{1+a^n}} $ converges or diverges. I want to show that $$\sum_{n=1}^\infty {\frac{a^n}{1+a^n}} $$ converges or diverges.
My guess is that it will depend heavily on the value of $a$.
So I would see the various cases of what $a$ could be.


*

*$a=0$ then its $\frac{0}{1}$ which converges absolutely to $0$.


But I am not sure in other cases. I think the other cases would be:


*

*$0<a<1$

*$-1<a<0$

*$a>1$

*$a<-1$


Is there any easier way to do it? If not, how would it look like for the other cases?
 A: Try to use the test for divergence:
$$\lim_{n \to \infty} \frac{a^n}{1 + a^n} \stackrel{LHR}{=} \lim_{n \to \infty} \frac{a^n \ln a}{a^n \ln a} = 1$$
for values of $a$ where the expression $\frac{a^n \ln a}{a^n \ln a}$ is defined ie $a > 0$

For $a=0$, you did that.

For $a < 0$, we can't use LHR, but...
$$\lim_{n \to \infty} \frac{a^n}{1 + a^n} = \lim_{n \to \infty} \frac{(-1)^n(-a)^n}{1 + (-1)^n (-a)^n} $$
Now, $\lim_{n \to \infty} (-1)^n$ does not exist. Hence, $\lim_{n \to \infty} \frac{a^n}{1 + a^n}$ dne.
A: Let $a$ be a real number. The series
$$
\sum_{n=1}^{\infty} \frac{a^{n}}{1 + a^{n}}
$$
converges (absolutely) if and only if $|a| < 1$.
Case 1: If $|a| < 1$, the geometric series
$$
\sum_{n=1}^{\infty} |a|^{n}
$$
converges. The reverse triangle inequality, and the fact that $|a|^{n} < 1$ for all $n \geq 1$, give
$$
\left|\frac{a^{n}}{1 + a^{n}}\right|
  = \frac{|a|^{n}}{|1 + a^{n}|}
  \leq \frac{|a|^{n}}{1 - |a|^{n}}.
\tag{1}
$$
Since $|a|^{n} \to 0$,
$$
\lim_{n \to \infty} \left|\frac{1}{|a|^{n}} \cdot \frac{|a|^{n}}{1 - |a|^{n}}\right|
  = \lim_{n \to \infty} \frac{1}{1 - |a|^{n}}
  = 1.
$$
Consequently, the series
$$
\sum_{n=1}^{\infty} \frac{|a|^{n}}{1 - |a|^{n}}
$$
converges by limit comparison with the geometric series. Finally, by (1) and ordinary comparison,
$$
\sum_{n=1}^{\infty} \frac{a^{n}}{1 + a^{n}}
$$
converges absolutely.
Case 2: If $1 < |a|$, then $|a|^{n} \to \infty$. The triangle inequality gives
$$
\left|\frac{a^{n}}{1 + a^{n}}\right|
  = \frac{|a|^{n}}{|1 + a^{n}|}
  \geq \frac{|a|^{n}}{1 + |a|^{n}} \to 1,
$$
so the series
$$
\sum_{n=1}^{\infty} \frac{a^{n}}{1 + a^{n}}
$$
diverges because its terms do not approach $0$.
Case 3: If $a = 1$, the terms are all $1/2$; if $a = -1$, the odd-degree terms are undefined. In either case, the series diverges.

The same arguments handle complex $a$ with a bit of additional work; one merely need note that if $|a| = 1$ and $a \neq \pm 1$, then
$$
\left|\frac{a^{n}}{1 + a^{n}}\right| = \left|\frac{1}{1 + a^{n}}\right| \geq \frac{1}{2}
$$
does not converge to $0$, so the series diverges.
A: $0< a<1$: compare with $\sum a^n $. Recall that $\sum_0^{N-1} a^n = \frac{1-a^N}{1-a}$ 
$a> 1$: note that $\frac{a^n}{1+a^n}\rightarrow 1$ as $n$ grows.
$a=1$ then $\frac{a^n}{1+a^n}= \frac{1}{2}$
A: If $a>1 $ $$\lim_{n\to +\infty}{\frac{a^n}{1+a^n}}=1$$ therefore the series diverges.
While if $-1<a<1$ $${\frac{a^n}{1+a^n}}\sim_{\infty}{\frac{a^n}{1}} $$ and this converges.
A: HINT: 
$$
\frac{a^n}{1+a^n}=\frac{a^n+1-1}{1+a^n}=1-\frac1{1+a^n}
$$
