If $|c_k| \leq k^2 $ does $\sum \frac{(-2)^k +c_k^3}{3^k }$ converge?
My approach, using the tools that we have learnt so far and trying to be as clear as I possibly can:
If an infinite series converges absolutely, it converges. We will apply the absolute value to the summand and apply the triangle inequality: $$\left|\frac{(-2)^k +c_k^3}{3^k } \right|=\frac{|(-2)^k +c_k^3|}{3^k } \leq\frac{|(-2)^k| +|c_k^3|}{3^k } = \frac{2^k +|c_k|^3}{3^k } $$ Now we can apply our hypothesis that$|c_k| \leq k^2$
$$\frac{2^k +|c_k|^3}{3^k }\leq \frac{2^k +k^6}{3^k } $$ certainly the exponential function grows faster than $k^6$ (this is something we discussed in class and studied) $$\frac{2^k +k^6}{3^k } \leq \frac{2^k +2^k}{3^k }= 2\frac{2^k}{3^k}=2\left(\frac{2}{3}\right)^k $$ So we can finally conclude that by comparison: $\sum \left|\frac{(-2)^k +c_k^3}{3^k }\right| \leq \sum 2\left(\frac{2}{3}\right)^k $
The majorant series is convergent, hence the absolute series is convergent by comparison. Since the series is absolutely convergent, it is also convergent. Conclusion: the series is convergent $\square$
This feels like a lot of work, I can probably do it in a better way, but this is the first thing that came to mind.
PS: this is the TU Eindhoven 2WA30 midterm practice test of 2014. If any fellow students are practising here. Welcome and good luck! remember, analysis is fun!
We have that
$$ \frac{(-2)^k -k^6}{3^k }\le \frac{(-2)^k +c_k^3}{3^k }\le \frac{(-2)^k +k^6}{3^k }$$
and since the LHS and RHS series converge also the given series converges.
The result can be easily shown indeed suppose that $\sum a_n$ and $\sum c_n$ converge and eventually
$$a_n\le b_n \le c_n \implies 0\le b_n-a_n \le c_n-a_n$$
then by comparison also $\sum (b_n-a_n)$ converges but
$$\sum b_n= \sum (a_n+(b_n-a_n)=\sum a_n +\sum (b_n-a_n)$$
and therefore also $\sum b_n$ converges since it is the sum of two convergent series.