Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$ 
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
  
  
*
  
*a) Show the monotony of the partial sums
  
*b) estimate upwards
  
*c) remember the geometric series (I do not know how to use that here.)
  


The following is what I have done so far: 
To show by induction: $a_{k+1} < a_k \forall k \in \mathbb N_0$
Induction start: $n=1$
$a_2= \frac{3^2}{5^2+1}=\frac{9}{25+1}=\frac{9}{26}=\frac{18}{52} < \frac{26}{52}=\frac{1}{2}=\frac{3}{6}=\frac{3^1}{5^1+1}=a_1$
Induction step: 
$$\begin{align}
a_{k+1}&<a_k \\
\equiv \frac{3^{k+1}}{5^{k+1}+1} &< \frac{3^k}{5^k+1} \\
\equiv \frac{3^{k+1}}{5^{k+1}} &< \frac{3^k}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+1}} &< \frac{3^{k+1}}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+2}} &< \frac{3^{k+1}}{5^{k+1}} \\
\equiv \frac{3^{k+2}}{5^{k+2}+1} &< \frac{3^k+1}{5^{k+1}+1} \\
\equiv a_{k+2} &< a_{k+1} \\
\end{align}$$
To Show: $|a_k|= a_k$
$$
\begin{align}
|a_k| &= |\frac{3^k}{5^k+1}| \\
      &= \frac{|3^k|}{|5^k+1|} \\
      &= \frac{3^{|k|}}{5^{|k|}+|1|} \\
      &= \frac{3^k}{5^k+1} \\
      &= a_k
\end{align}$$
Because of the induction I can conclude that the sequence $\lim_{k \to \infty}a_k$ becomes smaller and smaller. 
And because of $|a_k|=a_k$ are all values $\forall k \in \mathbb N$ positive.
$\Rightarrow $ The sequence  $a_k$ is monotically decreasing. 
$\Rightarrow $ The series $\sum_{k=1}^{\infty}a_k$ is monotically increasing.
$$
\begin{align}
a_k = \frac{3^k}{5^k+1} &< \frac{3^k}{5^k}  \\
                        &< \frac{3^k}{3^k}  \\
                        &= 1 \\
\end{align}$$
$$\lim_{k \to \infty} 1 = 1$$
And thus: $\exists N \in \mathbb N$, such that
$$|a_k| \le 1, \forall \quad k \ge N$$
Using the direct comparison test there can be concluded that $\sum_{k=1}^{\infty}{a_k}$ converges.

Question: Is my proof correct?


We know that a geometric series $\displaystyle \sum_{k=0}^{\infty} r^k$ converges $\iff$ $|r|<1$. 
Let $b_k=\left(\dfrac 35 \right)^k$. 
For every $k \ge 0$, we have $0<a_k \le b_k$. Thus if $\displaystyle \sum_{k=0}^{\infty} b_k$ converges, by the comparison test $\displaystyle \sum_{k=0}^{\infty}a_k$ converges too. And we know that $b_k$ converges because it is a geometric series with $|r|<1$.

A few comments about your work. You wrote  
$\dfrac{|3^k|}{|5^k+1|} =\dfrac{3^{|k|}}{5^{|k|}+|1|}$
But $|3^k|=3^{|k|}$ is in general not true. Consider that $|3^{-2}|=\dfrac 19$ , while $3^{|-2|}=9$. What should have been written is $|3^k|=3^{k}$, because for any $k$ we have $3^k>0$.
There is also a mistake in the denominator; you seem to think that $|a+b|=|a|+|b|$. But this is again not true in general. Consider $a=1, b=-1$. It is actually true that $|a+b| \le |a|+|b|$, with equality holding $\iff$ $a, b$ are both the same sign, or if both or one of them is zero.
You also wrote at the bottom that there is some $N$ such that when $k\ge N$, $a_k<1$, so therefore the series converges by the comparison test. But that is not at all what the comparison test says. Consider $c_k = \dfrac 1k$. There does exist an $N$ such that $k \ge N \implies c_k<1$ (namely $N=2$), but $\displaystyle \sum_{k=1}^{\infty} \dfrac 1k = 1 + \dfrac 12 + \dfrac 13 + \dfrac 14 + \cdots$ is the harmonic series, which famously diverges. 
