0
$\begingroup$

This question was asked in my msters exam for which I am preparing and I was unable to solve two options and so asking for help here.

Let $(u_n)_{n\geq1}$ be a sequence real numbers satisfying following:

(i) $(-1)^n u_n \geq 0$ for all $n>0$.

(ii) $|u_{n+1}|<\frac12|u_n|$ for all $n>12$.

Prove that $\sum_{n>12} u_n$ converges to a non-zero real number, and that if $|u_{n-1}| < \frac12|u_n|$ for all $1<n<12$, then $\sum_{n\geq 1} u_n$ is a negative real number.

(I am writing only the options I couldn't solve.)

I am not able to prove these two assertions and would be really thankful for help.

$\endgroup$
3
  • $\begingroup$ What part can't you solve ? Which convergence test might you use ? $\endgroup$ – Yves Daoust Nov 4 '20 at 14:11
  • $\begingroup$ "$| u_{n-1}| < |u_n|/2$ for all 1<n<12 ," Did you mean $u_{n+1}$ here? $\endgroup$ – zhw. Nov 4 '20 at 16:44
  • $\begingroup$ @zhw. No statement is written as it was in question! $\endgroup$ – Ben Nov 10 '20 at 14:20
3
+50
$\begingroup$

Why does no one do this problem? It is not that hard.
The first part is just the test for convergence. We just need to check if $|u_n|$ is decreasing to zero from a certain point $n$, which is obvious under your condition.
For more information on this test, you can take a look at this link :
https://en.wikipedia.org/wiki/Alternating_series_test .
The idea is simple, you can easily imitate if you are not allowed to use it in your exam.


For the **second part** , it is not true. I know what your professor was trying to imply in this problem. However, there is not enough condition for the second part to hold.
That is mainly because this problem provides no control over the largeness of $u_{12}$ (also no connection amongst $(u_n,n<12)$, $u_{12}$, $(u_n,n>12)$ )
So we can choose $u_{12}$ large enough to make the limit positive.
We should expect more of something like: $$ |u_{11}|> \dfrac{|u_{12}|}{2}> \frac{|u_{13}|}{4} $$

Full answer for part 1
For clarity, we define a new sequence $(a_n):= ( (-1)^{n}u_{n+12})$,clearly:

  • $(a_n)$ is a sequence of nonnegative numbers.
  • $a_n \ge \frac{1}{2} a_{n+1}$ hence it's a decreasing sequence. Furthermore, from that inequality, we get:
  • $0 \le a_{n+1} \le \dfrac{1}{2^n} a_1 \forall n>1$, thus:
    $(a_n)$ is a sequence decreasing to zero.
    Consider the following sequence of finite sums : $$S_n := \sum_{k=1}^n (-1)^ka_k$$

So we see that:

  • $S_{2n+1}=S_{2n} -a_{2n+1} \le S_{2n} $
  • $ S_{2n+1} -S_{2n-1}= a_{2n}-a_{2n+1} \ge 0 $( $(a_n)$ decreasing)
  • $ S_{2n+2}- S_{2n} = -a_{2n+1}+a_{2n+2 } \le 0$ (by the same reason) So $(S_{2n+1})$ is an increasing sequence and $(S_{2n}) $ is a decreasing sequence.
    Furthermore, $S_{2n+1}$ is bounded above by $S_2$ ( $S_{2n+1} \le S_{2n} \le S_2$ ) , $S_{2n}$ is bounded below by $S_1$ (by the same reasoning).
    So they must have limit, we note $\lim S_{2n+1}= L_1$ and $\lim S_{2n}= L_2$ , respectively.

As $\lim a_n =0 $ , we see that: $$ L_1-L_2= \lim S_{2n+1}- \lim S_{2n} =- \lim a_{2n+1}=0 $$ Thus $L_1=L_2=L$, from that we imply the convergence of $(S_n)$
By noticing that, $S_{n} = \sum_{k>12}^{12+n} u_n$ , we also derive the desired convergence.

On the other hand, as $S_{2n}$ decreasing, we have that: $$ L = \lim S_{2n} \le S_2 = a_{2}-a_1 <0$$ ( as $ a_2< \frac{1}{2}a_1 $ )
Thus $L $ is non zero. $\square $

$\endgroup$
2
  • $\begingroup$ I am unable to understand Part 1st . Kindly elaborate on that . $\endgroup$ – Ben Nov 16 '20 at 14:07
  • $\begingroup$ Sorry if I was not being too clear. I just updated the first part for you. $\endgroup$ – Paresseux Nguyen Nov 16 '20 at 14:49
2
$\begingroup$

Hint

A simple induction on the 2nd condition yields $$ {|u_n|<\left({1\over 2}\right)^{n-13}|u_{13}|\quad,\quad n\ge 14 \\ |u_n|>2^{n-1}|u_1|\quad,\quad 1<n<12 } $$ Now try to find the maximal and/or minimal values of $\sum_{i=1}^{12}u_n$ and $\sum_{i=13}^{\infty}u_n$ using the 1st condition to conclude the 2nd part of the question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.