How many terms need to be added to approximate $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}$ with an error less to $10^{-5}$
So since this is an alternating series we know that $$|R_n|=|S-S_n|≤a_{n+1}$$ And so we know then that $$-\frac{1}{(2n+2)!}≤\frac{1}{10^5}≤\frac{1}{(2n+1)!}$$ isn't it? And from here then $$10^5≤(2n+2)!$$ which I'm doubtful if this is even correct, and that one would solve to be $n≥3$, but my book says it is $n≥4$. Is there anything wrong in my reasoning?