Given that the series the summation from n=1 to infinity of [(-1)^(n+1)/√n is convergent, find a value of n for which the nth partial sum is guaranteed to approximate the sum of the series to two decimal places.
a) 39,999
b) 3,999
c) 399
d) 39
well, the terms alternate + and - so if we find an n where the n th term magnitude - the n-1 th term magnitude difference is less than .015 we are there
I am sorry but I don't get it. Please help me more with this one. Thanks
check the Remainder Theorem for alternating series
To find a value of n for which the nth partial sum guarantees an approximation of the sum of the series to two decimal places, we need to determine the accuracy of the partial sum.
The series in question is ∑[(-1)^(n+1)/√n] from n=1 to infinity, where (-1)^(n+1) alternates between -1 and 1.
To approximate the sum of the series, we can calculate the value of the nth partial sum. The nth partial sum of this series is given by S_n = ∑[(-1)^(k+1)/√k] from k=1 to n.
To guarantee an approximation to two decimal places, we need to find the smallest value of n for which the difference between S_n and the actual sum of the series is less than 0.01.
Since the series is convergent, we can use the Alternating Series Test. According to this test, if the absolute value of the terms in the series decreases and approaches zero as n increases, and the series satisfies the condition lim(n→∞) [(-1)^(n+1)/√n] = 0, then the series converges.
In this case, (-1)^(n+1)/√n is decreasing and approaches zero as n increases, so the series satisfies the Alternating Series Test and converges.
To find a suitable value of n, we can calculate the nth partial sum for each option and compare it with the actual sum of the series.
a) For n = 39,999:
Calculate the nth partial sum: S_n = ∑[(-1)^(k+1)/√k] from k=1 to 39,999.
Compare S_n with the actual sum.
b) For n = 3,999:
Calculate the nth partial sum: S_n = ∑[(-1)^(k+1)/√k] from k=1 to 3,999.
Compare S_n with the actual sum.
c) For n = 399:
Calculate the nth partial sum: S_n = ∑[(-1)^(k+1)/√k] from k=1 to 399.
Compare S_n with the actual sum.
d) For n = 39:
Calculate the nth partial sum: S_n = ∑[(-1)^(k+1)/√k] from k=1 to 39.
Compare S_n with the actual sum.
By comparing the calculated values of S_n with the actual sum of the series, we can determine which value of n guarantees the desired approximation to two decimal places.