# Find the smallest A satisfying the inequality

• MHB
• lfdahl
In summary, the purpose of finding the smallest A satisfying the inequality is to determine the minimum value of A that will make the inequality true. This can be done using mathematical techniques such as algebraic manipulation, graphing, or trial and error. It is possible for there to be more than one smallest A satisfying the inequality, and it is not always necessary to find the smallest A. Some common mistakes when finding the smallest A include not considering all possible solutions, making mathematical errors, and misinterpreting the inequality. It is important to carefully check the solution to ensure it satisfies the original inequality.
lfdahl
Gold Member
MHB
Let $a_1 = 1$, $a_2 = 1$ and $a_n = a_{n-1}+a_{n-2}$ for each $n > 2$. Find the smallest real number, $A$, satisfying

$\sum_{i = 1}^{k}\frac{1}{a_{i}a_{i+2}} \leq A$

for any natural number $k$.

lfdahl said:
Let $a_1 = 1$, $a_2 = 1$ and $a_n = a_{n-1}+a_{n-2}$ for each $n > 2$. Find the smallest real number, $A$, satisfying

$\sum_{i = 1}^{k}\frac{1}{a_{i}a_{i+2}} \leq A$

for any natural number $k$.

we have
$\frac{1}{a_{i}a_{i+2}}= \frac{1}{a_{i+1}}\frac{a_{i+1}}{a_{i}a_{i+2}}$
$= \frac{1}{a_{i+1}}\frac{a_{i+2}- a_i}{a_{i}a_{i+2}}$ (from given condition)
$= \frac{1}{a_{i+1}}(\frac{1}{a_i} - \frac{1} {a_{i+2}})$
above term is positive and telescopic sum so maximum sum is sum at infinite and adding we get
$\frac{1}{a_1a_2} + \frac{1}{a_2a_3}$ and se get putting the values $\frac{3}{2}$

we have
$\frac{1}{a_{i}a_{i+2}}= \frac{1}{a_{i+1}}\frac{a_{i+1}}{a_{i}a_{i+2}}$
$= \frac{1}{a_{i+1}}\frac{a_{i+2}- a_i}{a_{i}a_{i+2}}$ (from given condition)
$= \frac{1}{a_{i+1}}(\frac{1}{a_i} - \frac{1} {a_{i+2}})$
above term is positive and telescopic sum so maximum sum is sum at infinite and adding we get
$\frac{1}{a_1a_2} + \frac{1}{a_2a_3}$ and se get putting the values $\frac{3}{2}$

With your telescoping sum, I get:
$\sum_{i=1}^{k}\frac{1}{a_{i}a_{i+2}} = \sum_{i=1}^{k}\left ( \frac{1}{a_{i}a_{i+1}}-\frac{1}{a_{i+1}a_{i+2}} \right )\\ =\frac{1}{a_{1}a_{2}}-\frac{1}{a_2a_3}+\frac{1}{a_2a_3}-...+ \frac{1}{a_{k-1}a_k}-\frac{1}{a_ka_{k+1}}+\frac{1}{a_ka_{k+1}}-\frac{1}{a_{k+1}a_{k+2}} \\ = \frac{1}{a_{1}a_{2}}-\frac{1}{a_{k+1}a_{k+2}} = 1-\frac{1}{a_{k+1}a_{k+2}}.$

So the smallest possible $A$, that satisfies the inequality is:
$\lim_{k\rightarrow \infty }\left ( 1-\frac{1}{a_{k+1}a_{k+2}} \right ) = 1.$

lfdahl said:

With your telescoping sum, I get:
$\sum_{i=1}^{k}\frac{1}{a_{i}a_{i+2}} = \sum_{i=1}^{k}\left ( \frac{1}{a_{i}a_{i+1}}-\frac{1}{a_{i+1}a_{i+2}} \right )\\ =\frac{1}{a_{1}a_{2}}-\frac{1}{a_2a_3}+\frac{1}{a_2a_3}-...+ \frac{1}{a_{k-1}a_k}-\frac{1}{a_ka_{k+1}}+\frac{1}{a_ka_{k+1}}-\frac{1}{a_{k+1}a_{k+2}} \\ = \frac{1}{a_{1}a_{2}}-\frac{1}{a_{k+1}a_{k+2}} = 1-\frac{1}{a_{k+1}a_{k+2}}.$

So the smallest possible $A$, that satisfies the inequality is:
$\lim_{k\rightarrow \infty }\left ( 1-\frac{1}{a_{k+1}a_{k+2}} \right ) = 1.$

it is my mistake.
Your telescopic sum and hence limit is right.

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## 1. What is the purpose of finding the smallest A satisfying the inequality?

The purpose of finding the smallest A satisfying the inequality is to determine the minimum value of A that will make the inequality true. This is useful in solving mathematical equations and inequalities, as well as in real-world applications such as optimization problems.

## 2. How do I find the smallest A satisfying the inequality?

To find the smallest A satisfying the inequality, you will need to use mathematical techniques such as algebraic manipulation, graphing, or trial and error. The specific method will depend on the complexity of the inequality and the available information.

## 3. Can there be more than one smallest A satisfying the inequality?

Yes, it is possible for there to be more than one smallest A satisfying the inequality. This can happen when the inequality has multiple solutions or when there are multiple ways to express the inequality.

## 4. Is it necessary to find the smallest A satisfying the inequality?

No, it is not always necessary to find the smallest A satisfying the inequality. In some cases, finding any value of A that satisfies the inequality may be sufficient for the problem at hand. However, finding the smallest A can be helpful in understanding the behavior of the inequality and in finding the most efficient solution.

## 5. What are some common mistakes when finding the smallest A satisfying the inequality?

Some common mistakes when finding the smallest A satisfying the inequality include not considering all possible solutions, making errors in mathematical calculations, and misinterpreting the inequality. It is important to carefully check your work and ensure that the solution satisfies the original inequality.

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