Question

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{{10}^n}n}}} \) (error \(| < 0.000005\))

Short answer

Yes, the series is convergent and the number of terms we need to add is \(4\).

Step-by-step solution

Expanding the series

\(\begin{aligned}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{{10}^n}n!}}} &= \frac{{{{( - 1)}^0}}}{{{{10}^0} \cdot 0!}} + \frac{{{{( - 1)}^1}}}{{{{10}^1} \cdot 1!}} + \frac{{{{( - 1)}^2}}}{{{{10}^2} \cdot 2!}} + \frac{{{{( - 1)}^3}}}{{{{10}^3} \cdot 3!}} + ...\\\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{{10}^n}n!}}} &= 1 - \frac{1}{{10}} + \frac{1}{{{{10}^2} \cdot 2!}} - \frac{1}{{{{10}^3} \cdot 3!}} + ...\end{aligned}\)

These terms are alternatively positive and negative, so it is an alternating series

Alternating series test:

For the given series, we have

\({b_n} = \frac{1}{{{{10}^n}n!}}\)

Consider,\({b_{n + 1}} - {b_n}\)

\(\begin{aligned}{b_{n + 1}} - {b_n} &= \frac{1}{{{{10}^{n + 1}}(n + 1)!}} - \frac{1}{{{{10}^n}n!}}\\{b_{n + 1}} - {b_n} &= \frac{1}{{{{10}^n} \cdot 10(n + 1)n!}} - \frac{1}{{{{10}^n}n!}}\end{aligned}\)

\(\begin{aligned}{b_{n + 1}} - {b_n} &= \frac{{1 - 10(n + 1)}}{{{{10}^{n + 1}}(n + 1)!}}\\{b_{n + 1}} - {b_n} &= \frac{{9 - 10n}}{{{{10}^{n + 1}}(n + 1)!}} < 0{\rm{ for all }}n \in N\end{aligned}\)

Thus \({b_{n + 1}} - {b_n} < 0\) for all \(n \in N\)

Checking the limit condition

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left( {{b_n}} \right) &= \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{{{10}^n}n!}}} \right)\\\mathop {\lim }\limits_{n \to \infty } \left( {{b_n}} \right) &= 0\end{aligned}\)

Since\({b_n} = \frac{1}{{{{10}^n}n!}}\), satisfies the conditions that\({b_{n + 1}} \le {b_n}\)and

\(\mathop {\lim }\limits_{n \to \infty } \left( {{b_n}} \right) = 0\).

So, the series is convergent.

Checking the number of terms which we need to add in order to find the sum to the indicated accuracy.

\(\begin{aligned}S &= \frac{{{{( - 1)}^0}}}{{{{10}^0} \cdot 0!}} + \frac{{{{( - 1)}^1}}}{{{{10}^1} \cdot 1!}} + \frac{{{{( - 1)}^2}}}{{{{10}^2} \cdot 2!}} + \frac{{{{( - 1)}^3}}}{{{{10}^3} \cdot 3!}} + ...\\ &= 1 - \frac{1}{{10}} + \frac{1}{{200}} - \frac{1}{{6000}} + ...\end{aligned}\)

Note that,\(\begin{aligned}{b_5} &= \frac{1}{{240000}} < \frac{1}{{200000}}\\ &= 0.00005\end{aligned}\)and

\(\begin{aligned}{S_4} &= 1 - \frac{1}{{10}} + \frac{1}{{200}} - \frac{1}{{6000}}\\ &= 1 - 0.1 + 0.005 - 0.000167\end{aligned}\)

So,

\(\begin{aligned}|{R_4}| &= |S - {S_4}|{\rm{ }} \le {\rm{ }}{{\rm{b}}_{4 + 1}}\\|{R_4}| &= {{\rm{b}}_5} < 0.000005\end{aligned}\)

So,

\(|S - {S_4}| \le {\rm{65 < 0}}{\rm{.000005}}\)

Here, the given series gives specified accuracyupto \(4\) terms