(a)show that sigma(n=0,inf) x^n/n! converges for allx (b)deduce that lim(n to in
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Question
(a)show that sigma(n=0,inf) x^n/n! converges for allx
(b)deduce that lim(n to inf)x^n/n! for all x
Explanation / Answer
Ratio test: lim n-->8 |A(n + 1) / A(n)| If the limit < 1, the series converges. If the limit > 1, the series diverges. If the limit = 1, the test is inconclusive. Since your series also has x in it, we are looking for values of x that allow the limit to be < 1. Here, A(n) = x^n / n! So, A(n + 1) = x^(n + 1) / (n + 1)! And A(n + 1) / A(n) = A(n + 1) · 1 / A(n) So, lim n-->8 | A(n + 1) / A(n) | = lim n-->8 | x^(n + 1) / (n + 1)! · n! / x^n | We can cancel x^n from x^(n + 1) in the numerator and x^n in the denominator, and we can cancel n! from (n + 1)! in the denominator and n! in the numerator. This leaves us with lim n-->8 | x / (n + 1) | Since x does not change as n-->8, we have lim n-->8 | x / (n + 1) | = 0 Since 0 < 1 regardless of x, this series converges for all x. By the way, the series represents e^x. EDIT: Remember, n! = n· (n - 1) · (n - 2) · ...· 3 · 2 · 1 So, (n + 1)! = (n + 1) · (n) · (n - 1) · (n - 2) · ...· 3 · 2 · 1 Which also means (n + 1)! = (n + 1) · n! So the n! cancels with the (n + 1)!, leaving the first factor (n + 1). Hope that clarifies!Related Questions
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