TITLE: Remainders in Alternating Series / Remainders in general
QUESTION [5 upvotes]: So there's a little part I get stuck on when I'm trying to find the remainders, I'll post two simple problems with my work and my answer compared to the book answer.  Please help me out and tell me why what I am doing or where my thought process is wrong.
Determine how many terms of the convergent series must be summed to be sure that that remainder is less than $10^{-4}$
$$
\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}
$$
So when solving this 
$$
\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}
$$
So when solving this I first take the absolute value and set it as: $R_n \leq \frac{1}{2(n+1)+1}$
Now I say $10^{-4} \leq \frac{1}{2(n+1)+1}$ which means that $10^4 \geq 2(n+1)+1$. Next we have $n$ such that $\frac{10^4 -3}{2} \approx 4999$.  However, the book states it is $5000$.
I'm always off one number and I just don't get it, because it doesn't math algebraically.

REPLY [2 votes]: You want your inequality to be
$10^{-4} > \frac{1}{2n+1}$
because 
 you need the term to be 
smaller than the bound.
So
$2n+1 > 10000$
or
$n
> \frac{9999}{2}
= 4999.5
$
or
$n \ge 5000$.