What\'s the difference? Here they substituted a n+1 with a n But here they actua

Question

What's the difference?

Here they substituted an+1 with an

But here they actually used an+1

Given convergent series is pi squareroot 3/9 + ln 2/3 = sigma^infinity_k = 0 (-1)^k/3k + 1 Here a_k = 1/3k + 1 Since the series starts with k = 0, so (n + 1)^th term is a_n = 1/3n + 1 The Remainder R_n satisfies R_n lessthanorequalto a_n +1 R_n lessthanorequalto 1/3n + 1 Given convergent series is sigma^infinity_k = 0 (-1)^k/(2k +1)^3 Here a_k = 1/(2k + 1)^3 If absolute error less than 10^-3 Then R_n lessthanorequalto a_n + 1 10^3 2n > 7 n > 7/2 The above inequality is satisfied for n = 4.

Explanation / Answer

Because it's written there that the series starts with k=0.
That means for 2nd term k=1,
for 3rd term k=2,
for nth term k=n-1,
for (n+1)th term k=n.

Since ak=1/(3k+1), (n+1)th term is given given by k=n, i.e. an=1/(3n+1).

Whereas in the next question, the series starts with k=1. (n+1)th term is given by k=n+1.

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