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?


you have $\frac{3^k}{5^k+1} < \frac{3^k}{5^k}= \left(\frac{3}{5}\right)^k$ than it follows $$ \sum_{k=1}^{\infty}a_k  = \sum_{k=1}^{\infty} \frac{3^k}{5^k+1} \leq \sum_{k=1}^{\infty} \frac{3^k}{5^k} =  \sum_{k=1}^{\infty} \left(\frac{3}{5}\right)^k = \sum_{k=0}^{\infty} \left(\frac{3}{5}\right)^{k+1}=\frac{3}{5} \left( \frac{1}{1-\frac{3}{5} } \right) = \frac{3}{2}$$
So you showed that a bigger series converges and after the comparison test  you are done. 
