Prove ##(1/(a_1+1) ) + (1/(a_2+1) ) ....##

[giokrutoi]

Homework Statement

a₁=3 a_n+1=0.5(a_n²+1)
"n" is natural number

prove that (1/(a₁+1) ) + (1/(a₂+1) ) ... + (1/(a_n+1) ) <1/2
is true for every n

The Attempt at a Solution

could not figure out anything

 

[haruspex]

I would look for a simpler sequence, geometric say, which serves as a lower bound for the a_n for all n > some threshold N. Then it would be a matter of manually summing the 1/(a_n+1) up to N and adding the sum corresponding to the infinite bounding sequence after N.

 

[Delta2]

Not sure, maybe one trick would be to work with the sequence $b_{n+1}=0.5b_n, b_1=3$ or the sequence $b_{n+1}=0.5b_n^2$. If you can prove that the corresponding sum for $b_n$ is <1/2 then the corresponding sum for $a_n$ also satisfies <1/2 because it is $b_n+1<a_n+1 \rightarrow \frac{1}{a_n+1}<\frac{1}{b_n+1}$

 

[Delta2]

Ok my initial suggestion about $b_n$ don't work, try $b_{n+1}=3b_n, b_1=3$. You can prove that $b_n<a_n , n\geq 4$, and also you prove easily that $\sum\limits_{n=1}^{\infty}\frac{1}{b_n}=\frac{1}{2}$ and then using inequality $\frac{1}{a_n+1}<\frac{1}{b_n+1}<\frac{1}{b_n}$ for n>=4.

 

[haruspex]

Ok my initial suggestion about $b_n$ don't work, try $b_{n+1}=3b_n, b_1=3$. You can prove that $b_n<a_n , n\geq 4$, and also you prove easily that $\sum\limits_{n=1}^{\infty}\frac{1}{b_n}=\frac{1}{2}$ and then using inequality $\frac{1}{a_n+1}<\frac{1}{b_n+1}<\frac{1}{b_n}$ for n>=4.

I feel my hint in post #2 was enough to be going on with. Please let the student have time to work with one hint before laying it out further.

 

[giokrutoi]

I would look for a simpler sequence, geometric say, which serves as a lower bound for the a_n for all n > some threshold N. Then it would be a matter of manually summing the 1/(a_n+1) up to N and adding the sum corresponding to the infinite bounding sequence after N.

i could not find any rule for adding every number from N that i took as a threshold it is not geometric progression to add up every number and find limit of function

 

[giokrutoi]

Ok my initial suggestion about $b_n$ don't work, try $b_{n+1}=3b_n, b_1=3$. You can prove that $b_n<a_n , n\geq 4$, and also you prove easily that $\sum\limits_{n=1}^{\infty}\frac{1}{b_n}=\frac{1}{2}$ and then using inequality $\frac{1}{a_n+1}<\frac{1}{b_n+1}<\frac{1}{b_n}$ for n>=4.

thanks i guess it is right but may it be proof for sum of all 1/3^n this
-1 + 1 + 1/3 + 1/3^2 + 1/3^n =
-1 +(3^n+1 -1)/ 3-1) = 1-(1/3)^n/2

 

[giokrutoi]

I feel my hint in post #2 was enough to be going on with. Please let the student have time to work with one hint before laying it out further.

and what about this
2a_n+1 - 2 = a_n²- 1
1/a_n+1= 1/a_n-1 + 1 / a_n+1

 

[haruspex]

it is not geometric progression

I did not say it is. I said try to find some other sequence b_n, perhaps a geometric progression, and some number N, such that:

• For n>N, b_n>=a_n
• Σ₁^N1/(a_n+1) + Σ_N+11/(b_n+1) < 1/2

 

[giokrutoi]

I did not say it is. I said try to find some other sequence b_n, perhaps a geometric progression, and some number N, such that:

• For n>N, b_n>=a_n
• Σ₁^N1/(a_n+1) + Σ_N+11/(b_n+1) < 1/2

ok got it so the way that delta^2 offered

 

[haruspex]

ok got it so the way that delta^2 offered

Yes. But I have not checked whether Delta^2's specific sequence and starting point work, or if they do, whether they're the simplest choices.

 

[giokrutoi]

Yes. But I have not checked whether Delta^2's specific sequence and starting point work, or if they do, whether they're the simplest choices.

the second post of him is right and easy also

 

[Ray Vickson]

ok got it so the way that delta^2 offered

Bounding will need to be done VERY carefully, because the "1/2" limit is pretty strict.

If we use the exact recursion $a_1 = 3, a_{n+1} = \frac{1}{2} (a_n^2 + 1), n = 1,2, \ldots$ and compute exact values up to about $n = 12$, we see that the sequnece $a_n$ grows rapidly. Using Maple and exact arithmetic whenever it is practical, we get
$$\begin{array}{l}
a_{12} =
150069266781857262292843669672858603\\
4769448280147464726353368783536748476\\
47440046640744854967963947245498612021\\
477388413904375180451656270390552278692\\
1452055548775397792172060973375592551323900\\
39967096883676820619379899904933201414659759\\
5705864255753847966680994896463920097666188\\
3908504636929575700712046785222999071294587426\\
7583554677334026682716307498157604046253018642062\\
0448794191288742297786522568403222907220885\\
\doteq .1500692668 \times 10^{418}
\end{array} $$
More to the point are the values of
$$S_n = \sum_{j=1}^n \frac{1}{a_j+1} $$
for $n = 1,2, \ldots$.

To 833 digit accuracy we find $S_{12} = 0.5$ (that is, 0.5 followed by 832 zeros), but to 834 digits of precision we find that $S_{12}$ is a tiny bit less than 0.5; it prints as 0.499..., which is 0.4 followed by 833 '9's . Of course, values of $S_n$ for $n > 12$ will be a tiny bit larger, and so the bound of 1/2---if true--- will be very, very close.

Note added in edit: If
$$ S_n = \frac{M_n}{N_n} $$
for integers $M_n, N_n$ we find that $N_n = 2 M_n+2$, at least for $n$ from 1 to 12. Thus,
$$ S_n = \frac{1}{2} \frac{M_n}{M_n+1} $$
for $n = 1, 2, \ldots, 12$.

The $M_n$ grow rapidly with $n$; for example, $M_{12}$ is an 834-digit integer: $M_{12} \doteq 5.630196208 \times 10^{833}$. Thus
$$ S_{12} \doteq \frac{1}{2} \left( 1 - 1.776137035 \times 10^{-834} \right) . $$