How can I approximate the sum of series $\sum_{n=1}^{\infty} \frac{(-1)^n}{3^n n!}$ to four decimal places?
I thought i was supposed to evaluate the term until it is less than $\frac{1}{10^n}$ (here, in this case n = 4, since to four decimal places.
This gives me the fifth term when $\frac{1}{3^5 5!}$ < $\frac{1}{10^n}$. But the answer given in the answer sheet is sixth term, and says the fifth term affect to four decimal places. If this is true, what could I do to evaluate a series to nth decimal places, since my way of solving it doesn't necessarily work out for every case?
Thank you in advance for your kind responses 🙂
Best Answer
In order to get $k$ decimal places of accuracy after rounding, you need an error less than $0.5 \times 10^{-k} = 5 \times 10^{-k-1}$.
To see this, consider the question of how large a number can be and still be zero after rounding to four decimal places? $0.00009$ will round to $0.0001$, not zero. But anything less than $0.00005$ will round to zero.
The remainder estimate for alternating series says that the remainder is less than the first unsummed term. So you are looking for the first number $n$ such that $$ \left|\frac{(-1)^{n+1}}{3^{n+1}(n+1)!}\right| < 5 \times 10^{-5} = \frac{1}{2 \times 10^4} $$ Equivalently, you want $n$ such that $3^{n+1}(n+1)! > 20000$. Notice that $3^4 \times 4! = 1944$ and $3^5 \times 5! = 29160$. Since the fifth term is small enough, the sum of the first four terms has the desired accuracy.
Since we know the value of the infinite sum explicitly, we can verify: $$ \sum_{n=1}^4 \frac{(-1)^n}{3^nn!} = -\frac{551}{1944} = -0.283436\dots $$ while $$ \sum_{n=1}^\infty \frac{(-1)^n}{3^nn!} = e^{-1/3} -1 = -0.283468 \dots $$ We see agreement in the first four decimal places.