Question

Determine the sum of the series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^5}}}} \) correct to four decimals.

Short answer

The sum of the series is approximately 0.9721.

Step-by-step solution

Given data

The series is \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}}}{{{n^5}}}} \).

Concept of convergence and divergence sequence

If\({a_n}\)is a sequence and\(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, then the sequence\({a_n}\)is said to be converges otherwise it diverges.

Simplify the expression

Consider the series, \({s_n} = \frac{{{{( - 1)}^{n + 1}}}}{{{n^5}}}\).

Substitute 7 for \(n\) in \({s_n}\),

\(\begin{array}{l}{s_7} = \frac{{{{( - 1)}^{1 + 1}}}}{{{1^5}}} + \frac{{{{( - 1)}^{2 + 1}}}}{{{2^5}}} + \frac{{{{( - 1)}^{3 + 1}}}}{{{3^5}}} + \frac{{{{( - 1)}^{4 + 1}}}}{{{4^5}}} + \frac{{{{( - 1)}^{5 + 1}}}}{{{5^5}}} + \frac{{{{( - 1)}^{6 + 1}}}}{{{6^5}}} + \frac{{{{( - 1)}^{7 + 1}}}}{{{7^5}}}\\{s_7} = \frac{1}{1} - \frac{1}{{{2^5}}} + \frac{1}{{{3^5}}} - \frac{1}{{{4^5}}} + \frac{1}{{{5^5}}} - \frac{1}{{{6^5}}} + \frac{1}{{{7^5}}}\\{s_7} = \left( {1 + \frac{1}{{{3^5}}} + \frac{1}{{{5^5}}} + \frac{1}{{{7^5}}}} \right) - \left( {\frac{1}{{{2^5}}} + \frac{1}{{{4^5}}} + \frac{1}{{{6^5}}}} \right)\\{s_7} = 1.00449 - 0.03236\end{array}\)

That is, \({s_7} = 0.9721\).

Substitute 8 for \(n\) in \({s_n}\),

\(\begin{array}{l}{s_8} = \frac{{{{( - 1)}^{1 + 1}}}}{{{1^5}}} + \frac{{{{( - 1)}^{2 + 1}}}}{{{2^5}}} + \frac{{{{( - 1)}^{3 + 1}}}}{{{3^5}}} + \frac{{{{( - 1)}^{4 + 1}}}}{{{4^5}}} + \frac{{{{( - 1)}^{5 + 1}}}}{{{5^5}}} + \frac{{{{( - 1)}^{6 + 1}}}}{{{6^5}}} + \frac{{{{( - 1)}^{7 + 1}}}}{{{7^5}}} + \frac{{{{( - 1)}^{8 + 1}}}}{{{8^5}}}\\{s_8} = \frac{1}{1} - \frac{1}{{{2^5}}} + \frac{1}{{{3^5}}} - \frac{1}{{{4^5}}} + \frac{1}{{{5^5}}} - \frac{1}{{{6^5}}} + \frac{1}{{{7^5}}} + \frac{1}{{{8^5}}}\\{s_8} = \left( {1 + \frac{1}{{{3^5}}} + \frac{1}{{{5^5}}} + \frac{1}{{{7^5}}}} \right) - \left( {\frac{1}{{{2^5}}} + \frac{1}{{{4^5}}} + \frac{1}{{{6^5}}} + \frac{1}{{{8^5}}}} \right)\\{s_8} = 1.00449 - 0.03239\end{array}\)

Simplify further the above sum, \({s_8} \approx 0.9721\).

Since, \({s_7} = {s_8}\), it can be concluded that the sum of the series is approximately equal when \(n > 8\).

Therefore, the sum of the series is approximately equal to 0.9721.