Help with two limit problems I have to prove these limits by definition. Can somebody help?
$\lim_{n \to \infty}\left ( \frac{n^2+1}{4n^2+5} \right )=\frac{1}{4}$
$\lim_{n \to \infty}\left ( \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n} \right )=1$
This is what i have done so far:
For the first I got this and i don't know if its right and how to proceed: $\frac{\frac{-1}{\epsilon }-20}{16}<n^2$
For the second: $\frac{-2}{\epsilon+1} < 2^n$.
Thanks.
 A: For the first one:
$$\left | \frac{n^2+1}{4n^2+5} -\frac{1}{4}\right |=\left | \frac{4n^2+4-4n^2-5}{4(4n^2+5)}\right |= \frac{1}{4(4n^2+5)}<\epsilon \\\iff 4n^2+5>\frac{1}{4\epsilon}\iff 4n^2>\frac{1}{4\epsilon}-5\iff n^2>\frac{1}{16\epsilon}-\frac{5}{4}=\frac{1-5\epsilon}{16\epsilon} \\\iff n>\sqrt{\frac{1-5\epsilon}{16\epsilon} }. $$
Thus, for any $\epsilon >0$ ($\epsilon<1/5$) there exists $N>\sqrt{\frac{1-5\epsilon}{16\epsilon} }$ such that $$n\ge N \implies \left | \frac{n^2+1}{4n^2+5} -\frac{1}{4}\right |<\epsilon.$$
For the first one:
$$\left | \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}-1 \right|=\left|\frac{\frac12-\frac{1}{2^{n+1}}}{1-\frac12}-1\right|=\left|1-\frac{1}{2^n}-1\right|=\frac{1}{2^n}<\epsilon \\ \iff2^n>\frac{1}{\epsilon}\iff n>\log_2 \frac{1}{\epsilon}.$$
Thus for any $\epsilon >0$ there exists $N>\log_2 \frac{1}{\epsilon}$ such that $$n\ge N \implies \left | \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}-1 \right|<\epsilon.$$
Note that this is more or less what you have got. But you have forgotten the absolute value.
