MHB Jacob's question at Yahoo Answers (Alternating series approximation)

[Fernando Revilla]

Here is the question:

Approximate the sum of the series correct to four decimal places.
∞
∑ (-1)^n/(5^n*n!)
n=1

Here is a link to the question:

Approx. series help please? - Yahoo! AnswersI have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Jacob,

The Leibniz criterion for the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1}a_n$ says: if $a_n$ decrease monotonically to $0$ as a sequence of positive numbers when $n$ approaches infinity, i.e. then the alternating series converges. Moreover, let $S$ denote the sum of the series, then the partial sum $S_k =\sum_{n=1}^k (-1)^{n-1}a_n$ approximates $S$ with error bounded by the next omitted term: $\left | S_k - S \right|\leq a_{k+1}$. Then, $$a_{k+1}=\dfrac{1}{5^{k+1}(k+1)!}<0.0001 \Leftrightarrow 10\;000<5^{k+1}(k+1)!\qquad (*)$$ The smallest positive integer satisfying $(*)$ is $k=4$ so, $$S\approx \sum _{n=1}^4\frac{(-1)^n}{5^nn!}=\ldots$$ which approximate the sum of the series to four decimal places.