Question

Approximately the sum of series correct to four decimal places

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^7}}}} \)

Short answer

The sum of series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^7}}}} \)correct to four decimal places is \(0.9855\)

Step-by-step solution

Expand the series:

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^6}}}} = 1 - \frac{1}{{64}} + \frac{1}{{729}} - \frac{1}{{4096}} + ...\)

As the series contains the alternative positive and negative signs so this series is analternative series.

The standard form of Alternating series is as follows:

\(\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} {b_n} = {b_1} - {b_2} + {b_3} - {b_4} + {b_5} - {b_6} + ...{\rm{ }},{b_n} > 0\)

Alternating series test:

Compare the series \({b_n}\)with alternating series

\(\sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} {b_n}\)

Here the term\(\)is said to be convergent

If it satisfies the following two conditions.

(i) \(n\) for all\(n\)

(ii) \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\)

Differentiate the function

To show the sequence \(\left\{ {\frac{1}{{{n^6}}}} \right\}\)is decreasing take the related solution

\(f(x) = \frac{1}{{{x^6}}}\)for the sequence\(\left\{ {{b_n}} \right\}\)

Differentiate the function: \(f\left( x \right) = \frac{1}{{{x^6}}}\)

\(\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left( {\frac{1}{{{x^6}}}} \right)\)

\( = \frac{{ - 6}}{{{x^7}}}\) Since, \(\frac{d}{{dx}}\left( {{u^n}} \right) = n{u^{n - 1}}\)

Therefore, the derivative is f(x)=\(\frac{{ - 6}}{{^{x7}}}\)

As the values of \(x\)are positive, the inequality f(x)=\(\frac{{ - 6}}{X}\) for the variable x≥1

Thus, the function\(f(x) = \frac{1}{{{x^6}}}\)is decreasing on interval\((1,\infty )\).

Therefore, the sequence\(\left\{ {\frac{1}{{{n^6}}}} \right\}\)is decreasing sequence.

Hence, \({b_{n + 1}} < {b_n}\) for all \(n \ge 1\)

Finding the limit \({b_n}\)

\(\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{{n^6}}}} \right) = {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}} \right)^6} = 0\)

Thus the series\(s = \sum {{{( - 1)}^{n - 1}}} {b_n}\)satisfies the two condition of the alternating series test.

Therefore, by the alternating test, the series \(\sum\limits_{n - 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^6}}}} \)is convergent.

Alternating Series Estimation Theorem

Let \(S = \sum {{{( - 1)}^{n - 1}}} {b_n}\)be the sum of an alternating series,

Satisfies the following condition

(i) \(0 \le {b_{n + 1}} \le {b_n}\) and

(ii) \(\mathop {\lim }\limits_{n \to \infty } {\rm{ }}{{\rm{b}}_n} = 0\)

Then, \(|{R_n}| = |S - {S_n}| \le {b_{n + 1}}\)

Considering few terms of the series

\(\begin{aligned}S &= 1 - \frac{1}{{{2^6}}} + \frac{1}{{{3^6}}} - \frac{1}{{{4^6}}} + \frac{1}{{{5^6}}} - \frac{1}{{{6^6}}} + ...\\ &= 1 - \frac{1}{{64}} + \frac{1}{{729}} - \frac{1}{{4096}} + \frac{1}{{15625}} - \frac{1}{{46656}} + ...\\ &= 1 - 0.01562 + 0.001371 - 0.0002 + 0.0006 - 0.000002 + ...\end{aligned}\)

Notice the terms \({b_6}\)

\(\begin{aligned}{b_5} &= \frac{1}{{{6^6}}}\\ &= \frac{1}{{46656}} < \frac{1}{{20000}}\\ &= 0.00005\end{aligned}\)

To find the sum of the series correct to four decimal places set error less than\(0.00005\)

The sum of the\(5\)terms of the series is:

\(\begin{aligned}{S_5} &= \sum\limits_{n = 1}^\infty {\frac{1}{{{n^6}}}} \\ &= 1 - 0.01562 + 0.001371 - 0.0002 + 0.0006\\ &= 0.98557\end{aligned}\)

Final answer

\(|S - {S_5}| \le {b_6}\)

S - {S_5} < 0.00005

As the error is less than\(0.00005\), it does not affect the fourth decimal place of the sum

\({S_5} = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^6}}}} \)

Therefor, the sum of the series \(S\) correct to four decimal places is \(S = 0.98557\)