determine the values p for which the limit of $a_n$ converges I came across the problem below.
$0<a_{n+1}\leq a_n+1/n^p$ where $p$ is a real number and $1\leq n$.
For which values of $p$ does $<a_n>$ converge?
First I thought it would look somewhat like the p-series, and I could actually obtain
$a_{n+1} \leq a_1 + \sum_{k=2}^{n-1} k^{-p}$
but since $a_n$ is not monotone, I couldn't get further with this.
Can anyone help me with it? I am a very new to real analysis.
Thank you in advance.
 A: Define the sequence...
$${\alpha^p}_{n,c}={\alpha^p}_{n-1,c}+\frac{1}{(n-1)^p}\quad:\quad {\alpha^p}_{1,c}=c$$
As you have correctly inferred...
$${\alpha^p}_{n,c}=c+\sum_{k=1}^n k^{-p}$$
Now, define the sequence of sets $A^p_{n,c}=(0,{\alpha^p}_{n,c}]$.
The objective is now to find some subset $P\subseteq\Bbb{R}$ such that for all $p\in P$:

*

*$A^p_{n,c}$ is non-empty for all $n$ and $c>0$.


*$A^p_{n,c}$ remains bounded for all $n$, $c>0$


*$\lim_{n\to\infty}A^p_{n,c}\ne(0,\infty)$
This amounts to showing that ${\alpha^p}_{n,c},\lim_{n\to\infty}{\alpha^p}_{n,c}\in(0,\infty)$ for all $p\in P$, $n\in\Bbb{N}$, and $c>0$. Alternatively, you can work in the extended reals and have the condition ${\alpha^p}_{n,c}\in(0,\infty)$.

Note that...
$$\lim_{n\to\infty} A^p_{n,c}=(0,c+\zeta(p)]$$
...where $\zeta$ is the Riemann-zeta function whenever $p>1$. Since the Dirichlet series...
$$\zeta_n(s)=\sum_{k=1}^n k^{-s}$$
...only converges for $s>1$, we know that $P\subseteq(1,\infty)$. Furthermore, because $0<\zeta(s)<\infty$ for all $s>1$, we have that $A^p_{n,c}$ is non-empty and bounded for all $p>1$ - hence $P=(1,\infty)$.
Since every convergent sequence $a_n$ satisfying the inequality must also have $a_n\in A^p_{n,a_1}$ for all $n$, the convergence of $a_n$ implies $p\in(1,\infty)$.
This, together with DanielWainfleet's answer, shows that $a_n$ converges if and only if $p\in(1\infty)$.
A: *

*If $p\le 1:$ We cannot prove that $a_n$ converges because we have a counter-example: $a_1=1$ and $a_{n+1}=a_n+1/n\le a_n+1/n^p.$ And we cannot prove that $a_n$ diverges because we may have $a_n=a_1$ for all $n.$


*If $p>1:$ Then $a_n$ is bounded, as  $0<a_{n+1}\le a_1+\sum_{k=1}^n1/k^p<a_1+\sum_{k=1}^{\infty}1/k^p.$
A bounded real sequence that is not convergent must have two sub-sequences converging to values $r, s$ with $r<s.$
By contradiction, suppose $a_n$ does not converge. Let $f:\Bbb N \to \Bbb N$ and $g:\Bbb N\to \Bbb N$ be strictly increasing, with $\lim_{n\to \infty}a_{f(n)}=r<s=\lim_{n\to \infty}a_{g(n)}.$
Take $m\in \Bbb N$ such that $(s-r)/3>\sum_{k=m}^{\infty}1/k^p.$
Take $p\in \Bbb N$ with $f(p)\ge m$ and $|a_{f(p)}-r|<(s-r)/3.$ So $a_{f(p)}<r+(s-r)/3.$
Take $q\in \Bbb N$ with $g(q)>f(p)$ and $|a_{g(q)}-s|<(s-r)/3.$ So $a_{g(q)}>s-(s-r)/3.$
Now we have $$s-(s-r)/3<a_{g(q)}\le a_{f(p)}+\sum_{k=f(p)}^{g(p)-1}1/k^p\le$$ $$\le a_{f(p)}+\sum_{k=m}^{g(p)-1}1/k^p<$$ $$<a_{f(p)}+\sum_{k=m}^{\infty}1/k^p<$$ $$<a_{f(p)}+(s-r)/3<$$ $$<(r+(s-r)/3)+(s-r)/3=$$ $$=s-(s-r)/3$$ implying $s-(s-r)/3<s-(s-r)/3,$ which is absurd. So by contradiction, if $p>1$ then $a_n$ converges.
