Summation of a?n from n=1 to infinity=1-1/10+1/100-1/1000.......

Question

For the following alternating series,

Σ a_n = 1-1/10+1/100-1/1000+1/10000 .......

^(from n=1 to infinity)

how many terms do you have to go for your approximation (your partial sum) to be within 1e-08 from the convergent value of that series?

Answer 1

You can use remainder theorem for alternating series.

a_n = (-1/10)^n-1

Remainder for n terms, Rn < |a_n+1| = 1/10ⁿ ≤ 10^-8

n = 8

Therefore, you need to have the first 8 terms.

Answer 2

It's a geometric series since a_n = (-1/10)^n-1.

Recall that if |x| < 1, then 1 + x + x² + x³ + ... = 1 / (1 - x).

In this case we have x = -1/10 so the sum is

1 + (-1/10) + (-1/10)² + (-1/10)³ + ... = 1 / (1 - (-1/10)) = 1 / (1 + 1/10)

= 1 / (11/10) = 10/11.

The absolute error in using the nth partial sum (a₁ + a₂ + ... + a_n) is:

E_n = |a_n+1 + a_n+2 + a_n+3 ...| = |(-1/10)ⁿ + (-1/10)ⁿ⁺¹ + (-1/10)ⁿ⁺² ...|

= |(-1/10)ⁿ * [1 + (-1/10) + (-1/10)² + ...]|= |(-1/10)ⁿ(10/11)|

= 1 / (11*10^n-1).

Notice that 10^-n-1 < E_n < 10^-n.

Thus E₇ > 10e-8 but E₈ < 10e-8 and so you need to add up the first eight terms.