<s>[INST] 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.
 [/INST] For your first series $\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}$, apply the Divergence Test as you did: $$\lim_{x\to\infty}\frac{x^2}{x^2-2x+5}=1\implies\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}\space\text{diverges}$$
For your second series, $$\sum_{n=1}^{\infty}\frac{6\cdot2^{2n-1}}{3^n}=6\sum_{n=1}^{\infty}\frac{2^{2n-1}}{3^n}=6\sum_{n=1}^{\infty}\frac{2^{2n}2^{-1}}{3^n}=3\sum_{n=1}^{\infty}\frac{2^{2n}}{3^n}=3\sum_{n=1}^{\infty}\frac{4^n}{3^n}=3\sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^n$$
Thus we have a geometric series with common ratio $|r|=\frac{4}{3}>1$, hence the series diverges. 
For the third series, note that $e\approx2.7$ throughout, we have: $$\sum_{n=1}^{\infty}\frac{2^n+4^n}{e^n}=\sum_{n=1}^{\infty}\frac{2^n}{e^n}+\sum_{n=1}^{\infty}\frac{4^n}{e^n}=\sum_{n=1}^{\infty}\left(\frac{2}{e}\right)^n+\sum_{n=1}^{\infty}\left(\frac{4}{e}\right)^n$$
The first series is a convergent geometric series with $|r|=\frac{2}{e}<1$ and the second is a divergent geometric series with $|r|=\frac{4}{e}>1$. Then recall a series that is the sum of a convergent and divergent series is divergent. So the third series diverges. 
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