Function

Consider a sequence a{n} given by a{1} =1/3, a{n+1} =a{n} + a{n} ^2. Let S=\sum{i=2}^2008 1/a_{I}, then find the value of [S] ? (where [ ] represent greatest integer function)

Note by Abhishek Pal
7 years, 10 months ago

No vote yet
2 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Your answer is 5.

Compute the first few values

f(1)=13 f(1) = \frac{1}{3}

f(2)=49 f(2) = \frac{4}{9}

f(3)=5281 f(3) = \frac{52}{81}

f(4)=69166561 f(4) = \frac{6916}{6561}

f(5)=9320693243046721 f(5) = \frac{93206932}{43046721}

Observe

1f(2)+1f(3)+1f(4)+1f(5)=12159315923301733=B \frac{1}{f(2)}+\frac{1}{f(3)}+\frac{1}{f(4)}+\frac{1}{f(5)} = \frac{121593159}{23301733} = B

The exact decimal representation of B is not important - it can be shown that 5.2<B<5.3 5.2 < B < 5.3

Now lets try to put a bound on ∑n=620081f(n) \sum\limits _{n=6}^{2008} \frac{1}{f(n)}

Let's try to calculate f(6):

f(6)=126997849699225961853020188851841 f(6) = \frac{12699784969922596}{1853020188851841}

(Although the numbers are big this can all be done by hand in a reasonable amount of time)

It can be shown that f(6)>6 f(6) > 6

From this, we see that f(7)>62+6=42 f(7) > 6^2+6 = 42

What's more important is that for n>6n>6, f(n)f(n−1)>2 \frac{f(n)}{f(n-1)} > 2

And so, f(n−1)f(n)<12 \frac{f(n-1)}{f(n)}<\frac{1}{2}

We conclude:

1f(6)<16 \frac{1}{f(6)} < \frac{1}{6}

1f(7)<112 \frac{1}{f(7)} < \frac{1}{12}

1f(8)<124 \frac{1}{f(8)} < \frac{1}{24}

... \text{...}

1f(n+6)<12n∗6 \frac{1}{f(n+6)} < \frac{1}{2^n*6}

Now we have ∑n=620081f(n)=∑n=020021f(n+6)<∑n=0200216∗2n \sum\limits _{n=6}^{2008} \frac{1}{f(n)}=\sum\limits _{n=0}^{2002} \frac{1}{f(n+6)}<\sum\limits _{n=0}^{2002} \frac{1}{6 * 2^n}

We know that 1+12+14+18+...+12n<2 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\text{...}+\frac{1}{2^n}<2 for every positive integer n.

And so, ∑n=0200216∗2n<13 \sum\limits _{n=0}^{2002} \frac{1}{6 * 2^n}<\frac{1}{3}

And finally, ∑n=620081f(n)<13 \sum\limits _{n=6}^{2008} \frac{1}{f(n)} < \frac{1}{3}

Meaning that ∑n=220081f(n)<5.3+13<6 \sum\limits _{n=2}^{2008} \frac{1}{f(n)} < 5.3 + \frac{1}{3} < 6

The lower bound for ∑n=220081f(n) \sum\limits _{n=2}^{2008} \frac{1}{f(n)} remains 5.25.2

This is sufficient to show that your sum is indeed between 5 and 6, and so your solution is 5.

Ivan Stošić - 7 years, 10 months ago

according to me it is correct but according to book it is incorrect.............................the answer is.....2

Abhishek Pal - 7 years, 10 months ago
×

Problem Loading...

Note Loading...

Set Loading...