# how many terms do you have to go for your approximation (your partial sum) to be within 0.0001 from the convergent value of that series? For the following alternating series,∑ for n=1 to infinity of asubn = 1- (1/10) + (1/100) - (1/1000) + .........how many terms do you have to go for your approximation (your partial sum) to be within 0.0001 from the convergent value of that series? please show work so I can understand. Thank you!

This is a geometric alternating sereis with first term, a = 1 and the common ratio,r is `(-1/10)` .

Sum of first n terms is,

`S_n = a(1-r^n)/(1-r)`

`S_oo = a/(1-r)`

`S_n = 1(1-(-1/10)^n)/(1-(-1/10)) = (10/11)(1-(-1/10)^n)`

`S_oo=1/(1-(-1/10)) = 10/11 `

|S_oo-S_n| <=0.0001

|10/11 - (10/11)(1-(-1/10)^n)| <= 0.0001

10/11|1-(1-(-1/10)^n)|<=0.0001

|(-1/10)^n|<=0.00011

If...

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This is a geometric alternating sereis with first term, a = 1 and the common ratio,r is `(-1/10)` .

Sum of first n terms is,

`S_n = a(1-r^n)/(1-r)`

`S_oo = a/(1-r)`

`S_n = 1(1-(-1/10)^n)/(1-(-1/10)) = (10/11)(1-(-1/10)^n)`

`S_oo=1/(1-(-1/10)) = 10/11 `

|S_oo-S_n| <=0.0001

|10/11 - (10/11)(1-(-1/10)^n)| <= 0.0001

10/11|1-(1-(-1/10)^n)|<=0.0001

|(-1/10)^n|<=0.00011

If n is even then,

(-1/10)^n <=0.00011

(1/10)^n <= 0.00011

n log 0.1 <= log0.00011

n (-2.3025)<= -9.1150

n=> 3.95. Therefore n must be greater than or equal to 4.

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