I'm having trouble proving that this series converges.
$$\sum_{n=0}^{\infty}\frac{\cos\left(n\pi\right)}{3n!+1}$$
Also how can I determine the amount of terms to count together to reach a sum that deviates at max 0,01 from the actual sum of the series?
Any advice is greatly appreciated!
As Fred stated in the comments: $$|a_n|\leq b_n=\frac{1}{n!}.$$ Therefore since the sum $\sum_{n=0}^{\infty}b_n$ converges, the given sum converges as well.
For the amount of terms, needed for accuracy $0.01$, it is better to rewrite the sum as follows: $$\sum_{n=0}^{\infty}a_n=\sum Even\ n+\sum Odd\ n=\sum_{n=0}^\infty\frac{1}{3(2n)!}+\sum_{n=0}^\infty\frac{-1}{3(2n+1)!}=\sum_{n=0}^\infty\frac{2n+1-1}{3(2n+1)!}=\sum_{n=0}^\infty\frac{2n}{3(2n+1)!}=\sum_{n=0}^\infty a_n^*$$ Now all summands are positive, so the sum only increases. We need to examine the rate of change of the sum and of the summands, in order to be able to make a proper guess about $N$ - the number of summands needed to have the needed accuracy. These rates of change are: $$\Delta_\sum=a_n^*=\frac{2n}{3(2n+1)!}$$ $$\Delta_{a_n^*}=a_n^*-a_{n-1}^*=\frac{2n}{3(2n+1)!}-\frac{2n-2}{3(2n-1)!}=\frac{2}{3}\left[\frac{n}{(2n+1)!}-\frac{(n-1)(2n)(2n+1)}{(2n+1)!}\right]=\frac{2}{3}\frac{-4n^3+4n^2+3n}{(2n+1)!}$$ To be sure that the sum won't increrase with more than $0.01$ after $n=N$, we can look for the $n$, saisfying $\Delta_{a_n^*}\leq 0.001$. By guess and check one sees that the first such $n$ is $n=4$. But let's not forget that $a_n^*=a_{2n}+a_{2n+1}$ and: $$\sum_{n=0}^N a_n^*=\sum_{n=0}^{2N} a_n$$ so we have to sum up at most $2*4=8$ members of $\{a\}_n$ to get accuracy up to second decimal place.
However it should be noted that this is a very rough method and higher accuracies will require much less summands than what can be estimated with this method. To achieve accuracy $0.01$ are in fact needed only six summands.