Write a proof for this theorem: Let the sum from n=1 to infinity of a-sub n be a series. Then the limit as n approaches infinity of the absolute value of a-sub (n+1) divided by a-sub n equals one if and only if the limit as n approaches infinity of the nth root of the absolute value of a-sub n equals one.

Question
Asked Nov 7, 2019

Write a proof for this theorem: 

Let the sum from n=1 to infinity of a-sub n be a series. Then the limit as n approaches infinity of the absolute value of a-sub (n+1) divided by a-sub n equals one if and only if the limit as n approaches infinity of the nth root of the absolute value of a-sub n equals one. 

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Step 1

Given sum from n=1 to infinity of a-sub n be a series. Then

а,
=1 iff lim a, 1
lim
n+1
а,
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а, =1 iff lim a, 1 lim n+1 а,

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Step 2

Let use another sequence <bn

а,
а,
а-1
а,
а,
(1)
= a
then bb,b .b
а.
lim
n
n+1
1
а,
а,
lim
= 1
(using (1))
lim b,=1
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а, а, а-1 а, а, (1) = a then bb,b .b а. lim n n+1 1 а, а, lim = 1 (using (1)) lim b,=1

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Step 3

There exists a sequence <bn> such t...

lim b1
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lim b1

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