True or False? #2

Calculus Level 3

Which of the followings are true?

A. If a sequence a n a_n satisfies lim n a n = 0 , lim n k = 1 n a k \displaystyle \lim_{n\to\infty}a_n=0,~\lim_{n\to\infty} \sum_{k=1}^n a_k always has a value.

B. If a sequence a n a_n lets lim n k = 1 n a k \displaystyle \lim_{n\to\infty} \sum_{k=1}^n a_k to have a value, lim n a n = 0. \displaystyle \lim_{n\to\infty}a_n=0.

C. If a sequence a n a_n satisfies lim n a 2 n = α , \displaystyle \lim_{n\to\infty} a_{2n}=\alpha, then lim n a n = α . \displaystyle \lim_{n\to\infty} a_{n}=\alpha.


This problem is a part of <True or False?> series .

Only A and B Only C and A All of them Only B and C Only B Only A None of them Only C

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Boi (보이)
Jul 22, 2017

A. (counterexample)

a n = n + 1 n = 1 n + 1 + n a_n=\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}

lim n a n = 0 , \displaystyle \lim_{n\to\infty} a_n=0, but lim n k = 1 n a k \displaystyle \lim_{n\to\infty} \sum_{k=1}^n a_k doesn't have a value.

FALSE \therefore~\boxed{\text{FALSE}}


B.

If lim n k = 1 n a k \displaystyle \lim_{n\to\infty} \sum_{k=1}^n a_k has a value, then lim n k = 1 n + 1 a k \displaystyle \lim_{n\to\infty} \sum_{k=1}^{n+1} a_k also has the same value.

Therefore, lim n k = 1 n + 1 a k lim n k = 1 n a k = lim n a n = 0. \displaystyle \lim_{n\to\infty} \sum_{k=1}^{n+1} a_k-\lim_{n\to\infty} \sum_{k=1}^n a_k=\lim_{n\to\infty}a_n=0.

TRUE \therefore~\boxed{\text{TRUE}}


C. (counterexmple)

a n = ( 1 ) n . a_n=(-1)^n.

lim n a 2 n = 1 , \displaystyle \lim_{n\to\infty} a_{2n}=1, but lim n a n \displaystyle \lim_{n\to\infty} a_{n} doesn't have a value.

FALSE \therefore~\boxed{\text{FALSE}}


From above, only B is true.

I'd like to add that statement B is simply a restatement of the n th n^{\text{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 , 1 2 , 1 3 , 1 4 1 n } \{a_n\} = \{1, \dfrac 12, \dfrac 13, \dfrac 14 \ldots \dfrac 1n\} and its corresponding series n = 1 1 n \displaystyle \sum_{n \ = \ 1}^{\infty} \dfrac 1n .

Zach Abueg - 3 years, 10 months ago

Log in to reply

Yes, I know, but I was too lazy to prove n = 1 1 n \displaystyle \sum_{n \ = \ 1}^{\infty} \dfrac 1n does not have a value. Thanks for making that clear to everybody though! :D

Boi (보이) - 3 years, 10 months ago

Log in to reply

Of course, H.M.!

The simplest proof that the harmonic series diverges is the following:

n = 1 1 n = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + 1 7 + 1 8 + \displaystyle \sum_{n \ = \ 1}^{\infty} \frac 1n = 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \cdots

Now consider the following series, a slightly altered version of the harmonic series:

1 + 1 2 + ( 1 4 + 1 4 ) + ( 1 8 + 1 8 + 1 8 + 1 8 ) + = 1 + 1 2 + 1 2 + 1 2 + \displaystyle 1 + \frac 12 + \left(\frac 14 + \frac 14\right) + \left(\frac 18 + \frac 18 + \frac 18 + \frac 18\right) + \cdots \\ = 1 + \frac 12 + \frac 12 + \frac 12 + \cdots

Notice that the above series is clearly divergent; it is 1 + 1 + a geometric series with ratio r = 1 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?

Zach Abueg - 3 years, 10 months ago

Log in to reply

@Zach Abueg Those are some nice examples! It took me long enough to find out my counterexample. xD

Boi (보이) - 3 years, 10 months ago

Log in to reply

@Boi (보이) Haha no worries, man. Great concept problem!

Zach Abueg - 3 years, 10 months ago

Log in to reply

@Zach Abueg Thank you :)

Boi (보이) - 3 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...