Which of the followings are true?
A. If a sequence a n satisfies n → ∞ lim a n = 0 , n → ∞ lim k = 1 ∑ n a k always has a value.
B. If a sequence a n lets n → ∞ lim k = 1 ∑ n a k to have a value, n → ∞ lim a n = 0 .
C. If a sequence a n satisfies n → ∞ lim a 2 n = α , then n → ∞ lim a n = α .
This problem is a part of <True or False?> series .
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I'd like to add that statement B is simply a restatement of the n th term test for divergence, and statement A is its false converse. Another, very prominent counterexample to statement A is the sequence { a n } = { 1 , 2 1 , 3 1 , 4 1 … n 1 } and its corresponding series n = 1 ∑ ∞ n 1 .
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Yes, I know, but I was too lazy to prove n = 1 ∑ ∞ n 1 does not have a value. Thanks for making that clear to everybody though! :D
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Of course, H.M.!
The simplest proof that the harmonic series diverges is the following:
n = 1 ∑ ∞ n 1 = 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1 + 7 1 + 8 1 + ⋯
Now consider the following series, a slightly altered version of the harmonic series:
1 + 2 1 + ( 4 1 + 4 1 ) + ( 8 1 + 8 1 + 8 1 + 8 1 ) + ⋯ = 1 + 2 1 + 2 1 + 2 1 + ⋯
Notice that the above series is clearly divergent; it is 1 + a geometric series with ratio r = 1 . Since this series is greater than the harmonic series, then it must also diverge to infinity. QED
Here's a fantastic Brilliant wiki on the matter: If the limit of a sequence is 0, does the series converge?
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@Zach Abueg – Those are some nice examples! It took me long enough to find out my counterexample. xD
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@Boi (보이) – Haha no worries, man. Great concept problem!
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A. (counterexample)
a n = n + 1 − n = n + 1 + n 1
n → ∞ lim a n = 0 , but n → ∞ lim k = 1 ∑ n a k doesn't have a value.
∴ FALSE
B.
If n → ∞ lim k = 1 ∑ n a k has a value, then n → ∞ lim k = 1 ∑ n + 1 a k also has the same value.
Therefore, n → ∞ lim k = 1 ∑ n + 1 a k − n → ∞ lim k = 1 ∑ n a k = n → ∞ lim a n = 0 .
∴ TRUE
C. (counterexmple)
a n = ( − 1 ) n .
n → ∞ lim a 2 n = 1 , but n → ∞ lim a n doesn't have a value.
∴ FALSE
From above, only B is true.