Find the smallest A satisfying the inequality

[lfdahl]

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 $k$.

 

[kaliprasad]

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 $k$.

Spoiler

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 $\frac{3}{2}$

 

[lfdahl]

Spoiler

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 $\frac{3}{2}$

Hi, kaliprasad

Spoiler

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:
\[\lim_{k\rightarrow \infty }\left ( 1-\frac{1}{a_{k+1}a_{k+2}} \right ) = 1.\]

 

[kaliprasad]

Hi, kaliprasad

Spoiler

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.

Spoiler

Your telescopic sum and hence limit is right.