Show that $1/\zeta(2k) = \sum_{m \le K} \mu (m)/m^{2k} + O(1/K)$.
I have already proved that $1/\zeta(s) = \sum_{m=1}^{\infty} \mu (m)/m^s$. But how do I show that if $k\ge 1$, $1/\zeta (2k) = \sum_{m \le K} \mu (m)/m^{2k} + O(1/K)$. (Here $O$ denotes the big-$O$ notation.
Since you have proved that $1/\zeta (s) = \sum \limits_{n=1}^{\infty} \mu (n)/n^s$ you can see that for $s=2$ one gets $1/\zeta (2) = \sum \limits_{n=1}^{\infty} \mu (n)/n^2$ and therefore it remains to show that $$\sum \limits_{n>T}\mu (n)/n^2=O(1/T).$$ The last sum has absolute value $$|\sum_{n>T}\mu(n)n^{-2}|\leq \sum_{n>T}|\mu(n)|n^{-2}$$ owing to the triangle inequality. The function $\mu(n)$ only assume the values $1,-1$ and $0$ and therefore it satisfies $|\mu(n)|\leq 1$ for all $n$. This implies that the last sum is at most $$\sum_{n>T}n^{-2}.$$ Note that the function $f(t)=1/t^2$ for $t$ real in the interval $[n-1,n]$ takes its minimum at $t=n$ and hence the area enclosed by the interval $[n-1,n]$ and the line $y=(n)^{-2}$ is $\leq \int_{n-1}^{n}t^{-2} \mathrm{d}t$.Summing this for $n=[T],[T]+1,[T]+2,\cdots$ shows that $$\sum_{n>T}n^{-2}\leq \int_{t\geq T-1}t^{-2}\mathrm{d}t$$ whose value is $\frac{1}{T-1}\leq \frac{2}{T}$ for all $T\geq 2$. We have therefore shown that $$|\sum_{n>T}\mu(n)n^{-2}|\leq \frac{2}{T}$$ for all $T\geq 2$ and this shows that $$\sum_{n>T}\mu(n)n^{-2}=O(\frac{1}{T}).$$ There are other ways to show this but they all require a bit more knowledge, for example one can use $|\sum_{n\leq x}\mu(n)n^{-1}|\leq 1$ with partial summation.
EDIT:The original question only had $k=1$. My answer can be adopted verbatim to all cases where one has some complex number $s$ with real part greater than $1$ instead of $2k$. More specifically, denoting the real part of $s$ by $\sigma$ one will get $$1/\zeta(s)=\sum_{m\leq K}\mu(m)/m^s+O(\sigma/K^{\sigma-1})$$ where the implied constant in the error term is absolute. As a very special case, one gets for $k\geq 1$ that $$1/\zeta(2k)=\sum_{m\leq K}\mu(m)/m^{2k}+O(k/K^{2k-1})$$ with an absolute implied constant. This is done by replacing $s$ by $2k$ above. Ofcourse the last equation is a stronger version of what the edited question asks for since $$ 1/K^{2k-1}\leq 1/K \ \ \ \forall k\geq 1.$$ Also notice that the proof does not use anything about $\mu(n)$ except that it is a bounded function, it applies for example to the cases where one has any Dirichlet character in place of $\mu(n)$. However if the character is nontrivial one can say much more; namely one can get an error term $O(\sigma/K^{\sigma})$ in place of $O(\sigma/K^{\sigma-1})$, this for example shows that all nontrivial L-functions can be analytically continued in the region $\sigma>0$.