Let $\{a_n\}$ be any sequence of positive real numbers.

Suppose for all $n$, $a_{n+1}\le a_n + \frac1{n^p}$. Find all positive $p$ such that we can guarantee $\{a_n\}$ always converge.

For example, $p=2$ works since $a_{n+1}\le a_n + \frac1{n^2}< a_n + \frac1{n(n-1)}=a_n+\frac1{(n-1)}-\frac1{n}$, and so the sequence $b_n = a_n + \frac1{n-1}$ converges (monotone decreasing and bounded below by 0), which means $a_n$ converges.

Would the answer be difference if we change $\le$ to $<$ in the original question?


If $p\leqslant1$, the convergence of the sequence $(a_n)$ is not guaranteed and in fact the divergence may happen, as shown by the case when $a_n=\sum\limits_{k\leqslant n-1}\frac1{k^p}$.

For every $p\gt1$, one can modify the argument you explained when $p=2$, as follows. The series $\sum\limits_n\frac1{n^p}$ converges hence $r_n=\sum\limits_{k\geqslant n}\frac1{k^p}$ is finite for every $n\geqslant1$. Furthermore, $r_n=r_{n+1}+\frac1{n^p}$ hence $a_{n+1}+r_{n+1}\leqslant a_n+r_n$. By hypothesis, $a_n+r_n\gt0$ for every $n$ hence the sequence $(a_n+r_n)$ is nonincreasing and bounded below by $0$, in particular $(a_n+r_n)$ converges. Since $r_n\to0$, $(a_n)$ converges.

Would the answer be differen(t) if we change $\le$ to $<$ in the original question?

No, replace each $a_n$ by $\frac12a_n$.

  • $\begingroup$ But knowing $a_{n+1} - a_n < \frac1{n^p}$ doesn't mean $|a_{n+1} - a_n| < \frac1{n^p}$ (what if LHS is negative?). $\endgroup$ – user70520 Oct 12 '13 at 18:30
  • 1
    $\begingroup$ @user70520 You are right, sorry for the piece of nonsense and please see the revised version. $\endgroup$ – Did Oct 12 '13 at 21:52
  • $\begingroup$ Very nice! Thanks. $\endgroup$ – user70520 Oct 13 '13 at 0:06

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