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?

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?

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

Recall that if |x| < 1, then 1 + x + x^{2} + x^{3} + ... = 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 (a_{1} + a_{2} + ... + 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 E_{7} > 10e-8 but E_{8} < 10e-8 and so you need to add up the first eight terms.

You can use remainder theorem for alternating series.

a_{n} = (-1/10)^{n-1}

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

n = 8

Therefore, you need to have the first 8 terms.