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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

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    $\begingroup$ Ratio Test. Maybe first make things easier by observing that the $k$-th term is less than $\frac{3^k+k}{k!}$, indeed less than $\frac{3^k+3^k}{k!}$. $\endgroup$ – André Nicolas Nov 25 '15 at 22:51
  • $\begingroup$ Cheat answer: you know it's going to converge because "factorials grow really really quickly" (faster than exponentials but slower than double exponentials) but you still have to actually prove it, which is, in my opinion, the boring part $\endgroup$ – Zubin Mukerjee Nov 25 '15 at 22:57
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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.

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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.

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