Find the smallest number n of terms needed to obtain an approximation of the series ∞∑k=1 3/k^3, accurate to 10^-6

Answer 1

Using integral:

^∞∫₁3/k³dk = ⁿ∫₁3/k³ dk + ^∞∫_n3/k³dk

The last term is the remainder and it has to be ≤ 10^-6, so

lim b→∞ (3/-2)[1/k²]_n^b ≤10^-6

3/(2n²) ≤ 10^-6

n² ≥ (3/2) 10⁶

n ≥√(3/2) 10³ ≥ 1224.7 --> n = 1225

Answer 2

To find the smallest number of terms \( n \) needed to obtain an approximation of the series \( \sum_{k=1}^{\infty} \frac{3}{k^3} \) accurate to \( 10^{-6} \), we can use the concept of partial sums.

The series is convergent, so the sum of the first \( n \) terms, denoted as \( S_n \), can be used as an approximation to the infinite series.

The partial sum \( S_n \) is given by:

\[ S_n = \sum_{k=1}^{n} \frac{3}{k^3} \]

To determine the smallest \( n \) such that the difference between \( S_n \) and the infinite sum is less than \( 10^{-6} \), we can iterate through values of \( n \) until this condition is met:

\[ \left| \sum_{k=n+1}^{\infty} \frac{3}{k^3} \right| < 10^{-6} \]

Here's a sample Python code snippet to illustrate the approach:

epsilon = 1e-6 # Target accuracy

n = 1

partial_sum = 3.0 # First term

while True:

n += 1

term = 3.0 / (n ** 3)

partial_sum += term

# Check if the difference is less than epsilon

if term < epsilon:

break

print(f"The smallest number of terms needed is {n}")

This code iteratively adds terms until the difference between consecutive terms becomes smaller than the specified \( \epsilon \). The variable \( n \) represents the smallest number of terms needed for the desired accuracy.