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

For the following alternating series,

Σ an = 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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2 Answers

It's a geometric series since an = (-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 + ... + an) is:

En = |an+1 + an+2 + an+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*10n-1).

Notice that 10-n-1 < En < 10-n.

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

You can use remainder theorem for alternating series.

an = (-1/10)n-1

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

n = 8

Therefore, you need to have the first 8 terms.