Question

Approximately the sum of series correct to four decimal places

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n - 1}}{n^2}}}{{{{10}^n}}}} \)

Short answer

The sum of series correct to four decimal places

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}{n^2}}}{{{{10}^n}}}} \) is\(0.0676\)

Step-by-step solution

Expanding the series:

We have the series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}}{n^2}}}{{{{10}^n}}}} \)

We write first few terms of the series

\(\begin{aligned}S &= \frac{1}{{10}} - \frac{{{2^2}}}{{{{10}^2}}} - \frac{{{3^2}}}{{{{10}^3}}} - \frac{{{4^2}}}{{{{10}^4}}} - \frac{{{5^2}}}{{{{10}^5}}} - \frac{{{6^2}}}{{{{10}^6}}}\\S &= 0.1 - 0.04 - 0.009 - 0.0016 - 0.00025 - 0.000036 + 0.0000049...\end{aligned}\)

Sum of first six terms:

Since \({b_7} = 0.00000049\),this will not affect the fourth decimal

Place of the sum of series, so we have to take sum of first

Six terms only,\({S_6} = 0.067614\).

Therefore, we have,

\(S \approx 0.0676\), correct to four decimal place.