Find 10 Partial Sums of the Series: Σ (from n=1 to ∞) 16/(3^n)

To find the first 10 partial sums of the series Σ (from n=1 to ∞) 16/(3^n), we first identify that this is a geometric series.

A geometric series is a series of the form Σ (from n=0 to ∞) a*r^n, where a is the first term and r is the common ratio. In our case:

• a = 16/3 (when n=1)
• r = 1/3

The formula for the sum of the first N terms of a geometric series is:

S_N = a * (1 – r^N) / (1 – r)

Now, we will calculate the partial sums for N = 1, 2, …, 10:

1. S_1 = (16/3) * (1 - (1/3)^1) / (1 - 1/3) = 16/3 * (1 - 1/3) / (2/3) = 8.00000
2. S_2 = (16/3) * (1 - (1/3)^2) / (1 - 1/3) = 16/3 * (1 - 1/9) / (2/3) = 10.66667
3. S_3 = (16/3) * (1 - (1/3)^3) / (1 - 1/3) = 16/3 * (1 - 1/27) / (2/3) = 11.11111
4. S_4 = (16/3) * (1 - (1/3)^4) / (1 - 1/3) = 16/3 * (1 - 1/81) / (2/3) = 11.29630
5. S_5 = (16/3) * (1 - (1/3)^5) / (1 - 1/3) = 16/3 * (1 - 1/243) / (2/3) = 11.38889
6. S_6 = (16/3) * (1 - (1/3)^6) / (1 - 1/3) = 16/3 * (1 - 1/729) / (2/3) = 11.42857
7. S_7 = (16/3) * (1 - (1/3)^7) / (1 - 1/3) = 16/3 * (1 - 1/2187) / (2/3) = 11.44444
8. S_8 = (16/3) * (1 - (1/3)^8) / (1 - 1/3) = 16/3 * (1 - 1/6561) / (2/3) = 11.45455
9. S_9 = (16/3) * (1 - (1/3)^9) / (1 - 1/3) = 16/3 * (1 - 1/19683) / (2/3) = 11.46032
10. S_10 = (16/3) * (1 - (1/3)^10) / (1 - 1/3) = 16/3 * (1 - 1/59049) / (2/3) = 11.46429

Thus, the first ten partial sums of the series, rounded to five decimal places, are:

• S₁ = 8.00000
• S₂ = 10.66667
• S₃ = 11.11111
• S₄ = 11.29630
• S₅ = 11.38889
• S₆ = 11.42857
• S₇ = 11.44444
• S₈ = 11.45455
• S₉ = 11.46032
• S₁₀ = 11.46429