How to determine whether the series converges or diverges? I have the following sequence and I would like to determine if it converges or diverges. 
$$
\sum_{k=1}^\infty = \frac{3^k +k}{k! + 2} 
$$
How can I determine that? 
Thank you
 A: Everything is positive so convergence and divergence will not be change by rearrangement.  Also, since the monotonically increasing sequence of partial sums will converge if it has an upper bound, we need only upper bound your series.
$\frac{3^k+k}{k!+2} < \frac{3^k+k}{k!} = \frac{3^k}{k!} + \frac{k}{k!} = \frac{3^k}{k!} + \frac{1}{(k-1)!}$
$\sum_{k=1}^\infty \frac{3^k}{k!} = -1 + \sum_{k=0}^\infty \frac{3^k}{k!} = -1+\mathrm{e}^3$.
$\sum_{k=1}^\infty \frac{1}{(k-1)!} = \sum_{k=0}^\infty \frac{1}{k!} = \mathrm{e}$
So your series is upper bounded by $\mathrm{e}^3 + \mathrm{e} -1$.  Hence, it converges.
A: It's important to look at the terms of your sequence for large $k$'s. If you can bound the terms e.g. by a geometric sequence, like $a_k \leq C q^k$, with some $q<1$, you know that your sequence converges. Now, you know that the factorial grows faster than any exponential, so you can bound it e.g. by $C_1 4^k>k!$ for large enough $k$. Further, you can omit the additional terms $+k$ and $+2$ in the numerator and denominator (why?). So, in the end you can bound the term by something like $a_k \leq C_2 (3/4)^k$, which show you that the series will converge.
I left out some of the details, but I think you can work them out yourself. BTW, of course, you could replace the 4 in the argument above with any other real number strictly larger than 3. That just changes a little bit the constant involved.
