Show whether $\sum_{k=1}^{\infty} \frac{3^k+k^2}{4^k +1}$ converges or diverges

Question

Here's the problem to be solved:

Show whether $\sum_{k=1}^{\infty} \frac{3^k+k^2}{4^k +1}$ converges or diverges

My attempt

$\sum_{k=1}^{\infty} \frac{3^k+k^2}{4^k +1} \leq \sum_{k=1}^{\infty} \frac{3^k+k^2}{4^k} = \sum_{k=1}^{\infty} (3/4)^k + \sum_{k=1}^{\infty} \frac{k^2}{4^k}$

The first term in the last equality converges, however, I'm struggling to show whether $\sum_{k=1}^{\infty} \frac{k^2}{4^k}$ converges or diverges. My hypothesis is that it converges, since $4^k$ is growing much faster than $k^2$, so I somehow must show that it's less than or equal to something else.

I'd be glad if you could share any tips. Thanks.

Answer 1

For $k \ge 2$, we have $3^k \ge k^2$, and that's all you need.

Answer 2

Hint:

$$4^k = e ^{k\log 4} > \frac{k^4 (\log4)^4}{4!}$$

Answer 3

Hint : Think about $$\frac{1}{1-x} = \sum_{k=0}^\infty x^k$$ and its derivatives. Notice that $$\frac{d}{dx} \left(\frac{1}{1-x}\right) = \sum_{k=0}^\infty kx^{k-1}$$ then try multiplying both sides by $x$ and take another derivative and see what infinite sum we find.

This tells you it converges as well as its sum.

Answer 4

It is much simpler to use asymptotic analysis:

As $k^2=o(3^k)$, we have $3^k+k^2\sim_\infty3^k$. Similarly, $4^k+1\sim_\infty 4^k$. Therefore $$\frac{3^k+k^2}{4^k +1}\sim_\infty\biggl(\frac 34\biggr)^{\mkern-6mu k},$$which is a convergent geometric series.