Alternating series error bound, finding smallest n $$\sum_{n=1}^\infty \frac{(-1)^n(n^2)}{2^n}$$
Determine the smallest n guaranteeing that $$|R_n| < 0.01 $$
So I tried on this problem and use the formula $$|R_n| < a_{n+1}$$ and ended up with $$\frac{2^{n+1}}{(n+1)^2} > 100 $$ and this is where I stop because I have no clue how to find the smallest n, but I find out that 13 < n < 14 
 A: It should be $|R_n| \lt |a_{n+1}|$ because the $a$s can be negative, but that is not important.  You are correct that you want $\frac {2^{n+1}}{(n+1)^2} \gt 100$.  By whatever search technique you use (I would just make a spreadsheet and copy down for this) you find that $n+1=14$ gives $\frac {2^{14}}{14^2}=\frac {16384}{196} \lt 100$, but $n+1=15$ gives gives $\frac {2^{15}}{15^2}=\frac {32768}{225} \gt 100$ so you require $n \ge 14$
A: I believe you are on the right track with your last equation, although I do not think the +1 terms are appropriate.
My first inclination was to solve for when
(n^2)/(2^n) =0.01
Which is what you mostly have, except you've inverted it, and added the +1 terms, I assume in attempt to find the term right before the series achieves accuracy of 0.01. 
I do not believe this is correct, or at least it's more difficult than it has to be. 
Solving for accuracy = 0.01 will very likely not be an integer, as you indicated with your finding that the value of n is between 13 & 14. 
Using +1 will ruin this. 
My advice would be to use the setup I provided and fine-tune from there.
A: Just for your curiosity.
There is an explicit solution for $n$
$$\frac{2^{n+1}}{(n+1)^2} =k \implies n=-1-\frac{2 }{\log (2)}W_{-1}\left(-\frac{\log (2)}{2 \sqrt{k}}\right)$$ where appears the lower branch of Lambert function.
You can approximate it using the series expansion given in the linked page.
For $k=100$, this would give $n=13.3247$ and for $k=1000$, $n=17.3633$
