Subadditive lemma for $a_{m+n+2}leq a_n+a_m+g(m)$
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I have a sequence $a_n$ such that $a_{m+n+2}leq a_n+a_m+g(m)$ where $lim_{m to infty}frac{g(m)}{m}=0$ and $g(m)$ is a positive incresing function.
I need to show that $lim_{n to infty}frac{a_n}{n} =inf{frac{a_n}{n}}$.
Here's what I've came up so far:
Let there be a fixed $k>2$. For all $n>2$ we have that $n = mk+r+2$ with $0leq r <k$ so
$a_n = a_{mk+r+2}leq a_{mk}+a_r+g(r)leq a_{(m-1)k}+a_{k-2}+g(k-2)+a_r+g(r)leq ...leq$
$leq a_k+(m-1)a_{k-2}+(m-1)g(k-2)+a_r+g(r)$.
Now we divide everything by n.
$$frac{a_n}{n}leq frac{a_k}{n}+frac{(m-1)}{n}a_{k-2}+frac{(m-1)}{n}g(k-2)+frac{a_r+g(r)}{n}$$
taking $n to infty$
$limsup frac{a_n}{n}leq frac{a_{k-2}}{k}+frac{g(k-2)}{k} forall k>2$ . In particular $limsup frac{a_n}{n}leq infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} = infleft{frac{a_{k-2}}{k}right} $
assuming $infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} not in left{a_1+g(1) , frac{a(2)}{2}+frac{g(2)}{2}right}$ (which is true in the case I need, so even if this is not true in every case, for me it's enough) (last equality comes from the fact that $g(n) = o(n)$)
Now, if I had that $infleft{frac{a_{k-2}}{k}right}leq infleft{frac{a_{k}}{k}right}$ then I've proven what I wanted, as I'd have that $limsupleq inf implies limsup=liminf = lim$ but I can't reach this result.
Any hints on how to proceed or any other ideas for this proof altogether?
real-analysis sequences-and-series analysis convergence
asked Nov 19 at 18:01
Matheus barros castro
375110
add a comment |
up vote
0
down vote
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I have a sequence $a_n$ such that $a_{m+n+2}leq a_n+a_m+g(m)$ where $lim_{m to infty}frac{g(m)}{m}=0$ and $g(m)$ is a positive incresing function.
I need to show that $lim_{n to infty}frac{a_n}{n} =inf{frac{a_n}{n}}$.
Here's what I've came up so far:
Let there be a fixed $k>2$. For all $n>2$ we have that $n = mk+r+2$ with $0leq r <k$ so
$a_n = a_{mk+r+2}leq a_{mk}+a_r+g(r)leq a_{(m-1)k}+a_{k-2}+g(k-2)+a_r+g(r)leq ...leq$
$leq a_k+(m-1)a_{k-2}+(m-1)g(k-2)+a_r+g(r)$.
Now we divide everything by n.
$$frac{a_n}{n}leq frac{a_k}{n}+frac{(m-1)}{n}a_{k-2}+frac{(m-1)}{n}g(k-2)+frac{a_r+g(r)}{n}$$
taking $n to infty$
$limsup frac{a_n}{n}leq frac{a_{k-2}}{k}+frac{g(k-2)}{k} forall k>2$ . In particular $limsup frac{a_n}{n}leq infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} = infleft{frac{a_{k-2}}{k}right} $
assuming $infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} not in left{a_1+g(1) , frac{a(2)}{2}+frac{g(2)}{2}right}$ (which is true in the case I need, so even if this is not true in every case, for me it's enough) (last equality comes from the fact that $g(n) = o(n)$)
Now, if I had that $infleft{frac{a_{k-2}}{k}right}leq infleft{frac{a_{k}}{k}right}$ then I've proven what I wanted, as I'd have that $limsupleq inf implies limsup=liminf = lim$ but I can't reach this result.
Any hints on how to proceed or any other ideas for this proof altogether?
real-analysis sequences-and-series analysis convergence
asked Nov 19 at 18:01
Matheus barros castro
375110
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a sequence $a_n$ such that $a_{m+n+2}leq a_n+a_m+g(m)$ where $lim_{m to infty}frac{g(m)}{m}=0$ and $g(m)$ is a positive incresing function.
I need to show that $lim_{n to infty}frac{a_n}{n} =inf{frac{a_n}{n}}$.
Here's what I've came up so far:
Let there be a fixed $k>2$. For all $n>2$ we have that $n = mk+r+2$ with $0leq r <k$ so
$a_n = a_{mk+r+2}leq a_{mk}+a_r+g(r)leq a_{(m-1)k}+a_{k-2}+g(k-2)+a_r+g(r)leq ...leq$
$leq a_k+(m-1)a_{k-2}+(m-1)g(k-2)+a_r+g(r)$.
Now we divide everything by n.
$$frac{a_n}{n}leq frac{a_k}{n}+frac{(m-1)}{n}a_{k-2}+frac{(m-1)}{n}g(k-2)+frac{a_r+g(r)}{n}$$
taking $n to infty$
$limsup frac{a_n}{n}leq frac{a_{k-2}}{k}+frac{g(k-2)}{k} forall k>2$ . In particular $limsup frac{a_n}{n}leq infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} = infleft{frac{a_{k-2}}{k}right} $
assuming $infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} not in left{a_1+g(1) , frac{a(2)}{2}+frac{g(2)}{2}right}$ (which is true in the case I need, so even if this is not true in every case, for me it's enough) (last equality comes from the fact that $g(n) = o(n)$)
Now, if I had that $infleft{frac{a_{k-2}}{k}right}leq infleft{frac{a_{k}}{k}right}$ then I've proven what I wanted, as I'd have that $limsupleq inf implies limsup=liminf = lim$ but I can't reach this result.
Any hints on how to proceed or any other ideas for this proof altogether?
real-analysis sequences-and-series analysis convergence
asked Nov 19 at 18:01
Matheus barros castro
375110
I have a sequence $a_n$ such that $a_{m+n+2}leq a_n+a_m+g(m)$ where $lim_{m to infty}frac{g(m)}{m}=0$ and $g(m)$ is a positive incresing function.
I need to show that $lim_{n to infty}frac{a_n}{n} =inf{frac{a_n}{n}}$.
Here's what I've came up so far:
Let there be a fixed $k>2$. For all $n>2$ we have that $n = mk+r+2$ with $0leq r <k$ so
$a_n = a_{mk+r+2}leq a_{mk}+a_r+g(r)leq a_{(m-1)k}+a_{k-2}+g(k-2)+a_r+g(r)leq ...leq$
$leq a_k+(m-1)a_{k-2}+(m-1)g(k-2)+a_r+g(r)$.
Now we divide everything by n.
$$frac{a_n}{n}leq frac{a_k}{n}+frac{(m-1)}{n}a_{k-2}+frac{(m-1)}{n}g(k-2)+frac{a_r+g(r)}{n}$$
taking $n to infty$
$limsup frac{a_n}{n}leq frac{a_{k-2}}{k}+frac{g(k-2)}{k} forall k>2$ . In particular $limsup frac{a_n}{n}leq infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} = infleft{frac{a_{k-2}}{k}right} $
assuming $infleft{frac{a_{k-2}}{k}+frac{g(k-2)}{k}right} not in left{a_1+g(1) , frac{a(2)}{2}+frac{g(2)}{2}right}$ (which is true in the case I need, so even if this is not true in every case, for me it's enough) (last equality comes from the fact that $g(n) = o(n)$)
Now, if I had that $infleft{frac{a_{k-2}}{k}right}leq infleft{frac{a_{k}}{k}right}$ then I've proven what I wanted, as I'd have that $limsupleq inf implies limsup=liminf = lim$ but I can't reach this result.
Any hints on how to proceed or any other ideas for this proof altogether?
real-analysis sequences-and-series analysis convergence
real-analysis sequences-and-series analysis convergence
asked Nov 19 at 18:01
Matheus barros castro
375110
asked Nov 19 at 18:01
Matheus barros castro
375110
asked Nov 19 at 18:01
Matheus barros castro
375110
asked Nov 19 at 18:01
Matheus barros castro
375110
asked Nov 19 at 18:01
Matheus barros castro
375110
375110
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