To find the first 10 partial sums of the series given by:
\( S = \sum_{n=1}^{\infty} \frac{16}{3^n} \)
We can start by recognizing that this is a geometric series, where a (the first term) is \( \frac{16}{3} \) and the common ratio r is \( \frac{1}{3} \).
The formula for the n-th partial sum of a geometric series is:
\[ S_n = a \frac{1 – r^n}{1 – r} \]
Substituting our values in, we have:
\[ S_n = \frac{16}{3} \frac{1 – (\frac{1}{3})^n}{1 – \frac{1}{3}} = \frac{16}{3} \frac{1 – (\frac{1}{3})^n}{\frac{2}{3}} = 8(1 – (\frac{1}{3})^n) \]
This allows us to calculate the first 10 partial sums:
- \( n = 1: S_1 = 8(1 – (\frac{1}{3})^1) = 8(1 – \frac{1}{3}) = 8(\frac{2}{3}) = 5.33333 \)
- \( n = 2: S_2 = 8(1 – (\frac{1}{3})^2) = 8(1 – \frac{1}{9}) = 8(\frac{8}{9}) = 7.11111 \)
- \( n = 3: S_3 = 8(1 – (\frac{1}{3})^3) = 8(1 – \frac{1}{27}) = 8(\frac{26}{27}) = 7.70370 \)
- \( n = 4: S_4 = 8(1 – (\frac{1}{3})^4) = 8(1 – \frac{1}{81}) = 8(\frac{80}{81}) = 7.93580 \)
- \( n = 5: S_5 = 8(1 – (\frac{1}{3})^5) = 8(1 – \frac{1}{243}) = 8(\frac{242}{243}) = 7.99148 \)
- \( n = 6: S_6 = 8(1 – (\frac{1}{3})^6) = 8(1 – \frac{1}{729}) = 8(\frac{728}{729}) = 7.99725 \)
- \( n = 7: S_7 = 8(1 – (\frac{1}{3})^7) = 8(1 – \frac{1}{2187}) = 8(\frac{2186}{2187}) = 7.99911 \)
- \( n = 8: S_8 = 8(1 – (\frac{1}{3})^8) = 8(1 – \frac{1}{6561}) = 8(\frac{6560}{6561}) = 7.99984 \)
- \( n = 9: S_9 = 8(1 – (\frac{1}{3})^9) = 8(1 – \frac{1}{19683}) = 8(\frac{19682}{19683}) = 7.99998 \)
- \( n = 10: S_{10} = 8(1 – (\frac{1}{3})^{10}) = 8(1 – \frac{1}{59049}) = 8(\frac{59048}{59049}) = 7.99999 \)
Here are the rounded first 10 partial sums, each rounded to five decimal places:
- 5.33333
- 7.11111
- 7.70370
- 7.93580
- 7.99148
- 7.99725
- 7.99911
- 7.99984
- 7.99998
- 7.99999