Convergence of $\sum_{n=1}^\infty\frac{k^2}{k^2-2k+5}$, $\sum_{n=1}^\infty\frac{6\cdot 2^{2n-1}}{3^n}$, and $\sum_{n=1}^\infty\frac{2^n+4^n}{e^n}$ I have three series questions:


1) $$\sum_{n=1}^{\infty} \frac{k^2}{k^2 - 2k + 5}$$

I'm going to use this theorem:

So $$\lim_{x\to \infty} \frac{x^2}{x^2-2x+5}$$
So $$\lim_{x\to \infty} \frac{2x}{2x-2}$$
So $$\lim_{x\to \infty} \frac{2}{2} = 1$$
Is that right?

2) $$\sum_{n=1}^{\infty} \frac{6 \cdot  2^{2n-1}}{3^n}$$

I'm a tad stuck. I can get to here:
$$\sum_{n=1}^{\infty} \frac{2 \cdot 3 \cdot  2^{2n-1}}{3 \cdot 3^{n-1}}$$
$$\sum_{n=1}^{\infty} \frac{2^{2n}}{3^{n-1}}$$
Is this valid:
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2^{n}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2 \cdot 2^{n-1}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} 2^{n+1} \frac{2^{n-1}}{3^{n-1}}$$
So using this theorem:

it looks like this could be:
$$a_n = \frac{1}{1-r} = \frac{2^{n+1}}{\frac{1}{3}}$$
which diverges. Is this right? Is there a better way?

3) $$\sum_{n=1}^{\infty} \frac{2^n + 4^n}{e^n}$$

I have no idea where to even start.

In both cases you can use the necessary condition for convergence of a series:
If
$$\sum_{n=0}^\infty a_n$$
converges, then
$$a_n\to 0$$
(the same is true if the sum starts at some integer different from zero).
This implies that if $a_n\to 0$ is not true (whether because the limit is a non zero number, or infinity, or it doesn't exist), then
$$\sum_{n=0}^\infty a_n$$
does not converge.
In the first case, you have
$$\frac{n^2}{n^2-2n+5}=\frac{n^2}{n^2}\cdot\frac{1}{1-\tfrac2n+\tfrac5{n^2}}\to 1$$
as $n\to \infty$.
So
$$\sum_{n=0}^{\infty}\frac{n^2}{n^2-2n+5}$$
does not converge.
You can use a similar argument in the second example, since
$$\frac{6\cdot2^{2n-1}}{3^n}=\frac{6\cdot 2^{2n}\cdot 2^{-1}}{3^n}=\frac{6\cdot (2^2)^n}{2\cdot 3^n}=3\left(\frac43\right)^n\to \infty$$
as $n\to\infty$, since $\frac43>1$.
