Lovely summation
by cjquines0, Oct 6, 2016, 11:54 AM
For those of you who haven’t checked out Chen's new summation handout, do check it out. This problem is a practice problem from said handout. He has two other new handouts as well, which are both really good.
Cleaned up solution
False starts
Tidbit
Putnam 2011 wrote:
Let 
and 
be sequences of positive real numbers such that 
and 
for 
. Assume that the sequence 
is bounded. Prove that 
converges, and evaluate 
.
and 
be sequences of positive real numbers such that 
and 
for 
. Assume that the sequence 
is bounded. Prove that 
converges, and evaluate 
.Cleaned up solution
Note that the condition is equivalent to 
Also note that
This gives us
(The trailing terms in the sum tend to zero as
, making use of the fact that 
is bounded; this makes the cancellation in the telescoping sum possible.)
is equivalent to 
Also note that
This gives us
(The trailing terms in the sum tend to zero as
, making use of the fact that 
is bounded; this makes the cancellation in the telescoping sum possible.)False starts
The condition was iffy. I first rearranged it to 
, which didn't seem to be any helpful, but it at least gave me an idea for a quantitative estimate of 
given 
.
I tried messing with telescopes. I tried writing
as a function of 
and 
, and then using the given relation to see if 
has some clean values for 
and 
. But then I realized I couldn’t figure out how this would telescope with the third term.
I went back to the condition. Tried turning
, and then tried to find a relation between consecutive terms in . Tried taking reciprocals, tried getting the 
s to cancel out. I listed the first few terms of all in terms of and saw if the denominator was anything clean. Nothing.
I then tried a concrete example. I set
, giving 
. That would make 
I had no idea how to express that as a telescoping sum. I tried some false starts with this as well, like messing around with 
or something. No dice, once again. But I kept pushing on.
Went back to abstraction, trying
Still no dice, nothing was cancelling. I went back to the condition, and tried to express the ratio between terms of or of , trying to get a clean, neat ratio. No dice.
I finally got the crux idea of representing
That meant the numerator, 
, had to be 
. I go backwards and forwards, letting 
Then I tried solving in terms of 
. We had 
. We had 
or 
These looked like some of the relations I had earlier for! So I went back to the condition, back to . I wanted 
to be at the denominator, so I divided both sides by to get 
. That looked kinda weird and I was getting kinda worried. I placed the two fractions with denominators together because that would look cleaner, to get 
. I wanted a difference of the two terms to make the telescope, so I did 
. And I was staring at it, and I was like… wait, let’s add 
to both sides!
You can kinda guess what happened after that.
, which didn't seem to be any helpful, but it at least gave me an idea for a quantitative estimate of 
given 
.I tried messing with telescopes. I tried writing
as a function of 
and 
, and then using the given relation to see if 
has some clean values for 
and 
. But then I realized I couldn’t figure out how this would telescope with the third term.I went back to the condition. Tried turning
, and then tried to find a relation between consecutive terms in . Tried taking reciprocals, tried getting the 
s to cancel out. I listed the first few terms of all in terms of and saw if the denominator was anything clean. Nothing.I then tried a concrete example. I set
, giving 
. That would make 
I had no idea how to express that as a telescoping sum. I tried some false starts with this as well, like messing around with 
or something. No dice, once again. But I kept pushing on.Went back to abstraction, trying
Still no dice, nothing was cancelling. I went back to the condition, and tried to express the ratio between terms of or of , trying to get a clean, neat ratio. No dice.I finally got the crux idea of representing
That meant the numerator, 
, had to be 
. I go backwards and forwards, letting 
Then I tried solving in terms of 
. We had 
. We had 
or 
These looked like some of the relations I had earlier for
! So I went back to the condition, back to . I wanted 
to be at the denominator, so I divided both sides by to get 
. That looked kinda weird and I was getting kinda worried. I placed the two fractions with denominators together because that would look cleaner, to get 
. I wanted a difference of the two terms to make the telescope, so I did 
. And I was staring at it, and I was like… wait, let’s add 
to both sides!You can kinda guess what happened after that.
Tidbit
Hello again, dearest readers! I’m sure my writing is a far better than SSD’s, and that my post is a breath of fresh air to everyone (read: no one) who follows this blog.
I picked up olympiad math after a two-week hiatus. It was a very busy hiatus, filled with lots of schoolwork that filled my evenings, side projects that filled my school days, and Magic the Gathering in my free time (gasp). But finally, I have shambled back to the habit of doing math.
The side project occupying the largest amount of time is this enrichment program called (oddly enough) PRIME. It’s a program I handle that meets every Wednesdays and Fridays after class, and we discuss roughly AIME-level contest math. It’s pretty intensive, because we’re training for the upcoming first round of our national olympiad. Hopefully we can get a significant number of people to second round. I’ll be uploading the handouts I’ve used to my website once I edit out the errors and make an answer key.
Speaking of which, there are two math contests coming up in October over here. One is a fast-paced oral competition on October 19, which we are anticipating to be very difficult to qualify to final round of. The second is the said first round of our national olympiad on October 22. I’m hyped, because these will be the first two math contests I’ll join this school year, and I’ve been itching for some action.
Our semestral break will start on October 22, and it sucks because classes will resume November 2, with final exams on November 3 and 4. It’s too long of a break. I wanted a shorter one.
Enough rambling. I’ll post a bit more frequently. Bye for now.
I picked up olympiad math after a two-week hiatus. It was a very busy hiatus, filled with lots of schoolwork that filled my evenings, side projects that filled my school days, and Magic the Gathering in my free time (gasp). But finally, I have shambled back to the habit of doing math.
The side project occupying the largest amount of time is this enrichment program called (oddly enough) PRIME. It’s a program I handle that meets every Wednesdays and Fridays after class, and we discuss roughly AIME-level contest math. It’s pretty intensive, because we’re training for the upcoming first round of our national olympiad. Hopefully we can get a significant number of people to second round. I’ll be uploading the handouts I’ve used to my website once I edit out the errors and make an answer key.
Speaking of which, there are two math contests coming up in October over here. One is a fast-paced oral competition on October 19, which we are anticipating to be very difficult to qualify to final round of. The second is the said first round of our national olympiad on October 22. I’m hyped, because these will be the first two math contests I’ll join this school year, and I’ve been itching for some action.
Our semestral break will start on October 22, and it sucks because classes will resume November 2, with final exams on November 3 and 4. It’s too long of a break. I wanted a shorter one.
Enough rambling. I’ll post a bit more frequently. Bye for now.
so that ![\[S=\sum_{n=1}^{\infty} \left [ \frac{f(n)}{a_1 \cdots a_n}-\frac{f(n+1)}{a_1 \cdots a_{n+1}} \right ]. \]](http://latex.artofproblemsolving.com/9/2/c/92c4cb6e5b7880970413db0ba77639cf00d94f9f.png)
![\[=\sum_{n=1}^{\infty} \left [ \frac{f(n)}{a_1 \cdots a_n}-\frac{\frac{f(n+1)}{a_{n+1}}}{a_1 \cdots a_{n}} \right ]. \]](http://latex.artofproblemsolving.com/b/7/d/b7dcf41c4817efaa24e28d6495e0d086cd8ae598.png)
The 
in the denominator won't do, so we manipulate the given to be ![\[\frac{1}{a_{n+1}}=\frac{b_n}{b_{n+1}+2}. \]](http://latex.artofproblemsolving.com/c/b/3/cb3a73b5e62ff4948808581591d2f95df1181b6d.png)
![\[=\sum_{n=1}^{\infty} \left [ \frac{f(n)}{a_1 \cdots a_n}-\frac{f(n+1) \cdot \frac{b_n}{b_{n+1}+2}}{a_1 \cdots a_{n}} \right ].\]](http://latex.artofproblemsolving.com/c/f/e/cfe2990ddb7f95a809b73f62871ab41e38a061e9.png)
to get rid of the ugly denominator, and by now the sum is dead: ![\[ = \sum_{n=1}^{\infty} \left [ \frac{b_n+2}{a_1 \cdots a_n}-\frac{b_n}{a_1 \cdots a_{n}} \right ] \]](http://latex.artofproblemsolving.com/d/f/8/df8599c8b9b6072387f44104f302a7be74ddde0a.png)
![\[ = \frac{2}{a_1 \cdots a_n} \]](http://latex.artofproblemsolving.com/0/2/8/02891d02c304e3b73f9907629a13ff23046fe634.png)
and the answer follows.
May 2017
April 2017