# Proving that $\sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}$ converges by the comparison test

I would like to prove that the following series converges:$$\sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}$$ by comparing it with a series that I already know converges. One such series could be the geometric series $$\sum_{k=1}^\infty \frac{2^{k}}{3^{k}}$$. Since I know that $$\sum_{k=1}^\infty \frac{2^{k}}{3^{k}}$$ converges, the only thing left to prove is that

$$L = \lim_{k \to \infty}\frac{a_k}{b_k} < +\infty ,$$where $$a_k = \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}$$ and $$b_k = \frac{2^{k}}{3^{k}}$$. However, when I try to prove this I get the following problem:

$$\lim_{k \to \infty}\frac{3^{k}(k^{8}+2^{k})}{2^{k}(3^{k} - 2^{k})}$$

And I don't know how to solve this limit problem, and show that the above limit is less than $$+\infty$$ so I'm kinda stuck and would appreciate any help!

• The dominant term in both numerator and denominator is $6^k$. I'd cancel that to get $$\frac{1+2^{-k}k^8}{1-(2/3)^k}$$ Jul 27, 2020 at 14:30

There are two well-known comparison tests. The first is the Limit Comparison Test: roughly, it says that if $$a_k,b_k$$ are positive sequences with $$\lim a_k/b_k = L>0$$, then $$\sum a_k$$ and $$\sum b_k$$ converge or diverge together. The second is the Direct Comparison Test; roughly, it says that if $$0\leq a_k \leq c_k$$ and $$\sum c_k$$ converges, so does $$\sum a_k$$; likewise, if $$0\leq d_k\leq b_k$$ and $$\sum d_k$$ diverges, so does $$\sum b_k$$. You can use either test here.

For the LCT, you've correctly identified a good candidate for $$b_k$$. To compute the limit, try dividing by $$6^k$$ and then using $$\lim_{k\to\infty} k^n/ r^k = 0$$ if $$r>1$$ and $$n\in \mathbb{R}$$: $$\lim_{k\to \infty}\frac{3^k(2^k+k^8)}{2^k(3^k-2^k)} = \lim_{k\to \infty}\frac{1+k^8/2^k}{1-(2/3)^k}= \frac{1+0}{1-0}=1$$Since $$\sum b_k$$ converges, so does $$\sum a_k$$.

For the DCT, let's throw away some stuff to make a good comparison. Observe that $$k^8 < 10^{20}\cdot 2^k$$ for $$k\geq 1$$. Likewise, $$(5/2)^k /3<3^k - 2^k$$ for $$k\geq 1$$. Thus $$0 < \frac{2^k +k^8}{3^k-2^k} < \frac{(1+10^{20}) 2^k}{3^k-2^k} < \frac{(1+10^{20})2^k}{(5/2)^k / 3} = 3(1+10^{20}) \left(\frac{4}{5}\right)^k$$These are the terms of a convergent geometric series. So the original series converges.

• Thanks for going in depth. I've gone through your comment and understand most of it, there is just one thing i don't understand, and it might be a stupid question. But how do you see/prove that $k^8 < 10^{20}\cdot 2^k$? Jul 28, 2020 at 8:48
• Eventually $2^k>k^8$, but not so for small values of $k$. I chose the constant to be big enough to give some breathing room. To prove it, you could take logs and show the derivative of RHS is greater. Jul 28, 2020 at 14:15

$$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}= \left(\frac{2}{3}\right)^{k} \cdot \frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}}$$ As $$\frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}} \to 1$$, then we can say, that $$\exists N \in \mathbb{N}$$ such that for $$k>N$$ holds $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \frac{3}{2}$$, so $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \left(\frac{2}{3}\right)^{k-1}$$ for $$k>N$$.