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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!

Integrand
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Fosofof
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    The dominant term in both numerator and denominator is $6^k$. I'd cancel that to get $$\frac{1+2^{-k}k^8}{1-(2/3)^k}$$ – Angina Seng Jul 27 '20 at 14:30

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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.

Integrand
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  • 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 $? – Fosofof Jul 28 '20 at 08:48
  • 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. – Integrand Jul 28 '20 at 14:15
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$$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}= \left(\frac{2}{3}\right)^{k} \cdot \frac{1+ \frac{k^8}{2^{k}} }{1-\left( \frac{2}{3}\right)^{k}}$$ 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 $k>N$.

zkutch
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