Convergence of $na_{n+1}-\sum_{k=1}^n a_k$

Question

I am given an increasing sequence $\{a_n\}_{n\ge 1}$ such that $0< a_n\le 1,\ n\ge 1$. Can we say that the sequence $$b_n:=na_{n+1}-\sum_{k=1}^n a_k,\quad n\ge 1$$ converges? And if we can then are there conditions on $a_n$ for that? Thank you.

score 4 · Answer 1 · edited Apr 13 '17 at 12:20

Here is a partial answer that gives an upper bound on $b_n$, together with a counterexample to convergence in general. Since $$\frac{b_n}{n} = a_{n + 1} - \frac 1 n \sum_{k = 1}^n a_k$$

it follows that $b_n/n$ gives the difference between $a_{n + 1}$ and the average of the first $n$ terms of the sequence. Since $\{a_n\}$ is an increasing sequence, this limit converges to $0$; see, e.g. here. This gives a bound on the growth of $b_n$.

However, it's not true that $\{b_n\}$ must converge; consider a slowly increasing sequence, such as

$$a_k = 1 - \frac 1 k$$

Then $$\sum_{k = 1}^n a_k =n - H_n$$

with $H_n$ being the $n$th harmonic number. In this case, we have

$$n a_{n + 1} - \sum_{k = 1}^n a_k = n - \frac{n}{n + 1} - (n - H_n) = H_n - \frac{n}{n + 1}$$

which diverges to infinity. Note that it has logarithmic growth, in accordance with the bound above.

In some sense, the averages lag behind the current term, due to the slow convergence of the sequence (or the slow divergence of the partial sums $\sum (1 - a_n)$; faster convergence can guarantee that the limit exists. For example, if we choose $a_n = 1-\frac{1}{2^n}$, the corresponding sequence $\{b_n\}$ does converge.

score 4 · Answer 2 · answered May 18 '14 at 01:18

Since $\{ a_{n} \}$ is monotone increasing and bounded, it converges to a limit $\alpha$. Then it is easy to check that

$$ \lim_{n\to\infty}b_{n} = \sum_{n=1}^{\infty} (\alpha - a_{n}). $$

(Of course, we are not excluding the possibility that this diverges to $\infty$.) Since $c_{n} = \alpha - a_{n}$ can be any decreasing sequence having values in $[0, \alpha]$, there seems no easy criterion for the convergence of $\{ b_{n} \}$.

score 2 · Answer 3 · answered May 18 '14 at 00:02

Answer above confirms that $(b_n)$ can be both convergent or divergent. As for a precise characterisation of convergence of $(b_n)$ in terms on $(a_n)$ it might be useful to look at rates of Cauchy convergence, since the relationship between the two sequences is seen most clearly in terms of differences i.e. $$(b_{n+1}-b_n)=(n+1)(a_{n+1}-a_n).$$ It could be that one could prove something like "$(b_n)$ converges iff $(a_n)$ has rate of Cauchy convergence $\mathcal{O}(f(n))$" for some reasonably simple function $f(n)$.

Sorry if that sounds a bit cryptic: I haven't the energy right now to think any harder! Maybe it's much easier than that.

Convergence of $na_{n+1}-\sum_{k=1}^n a_k$

3 Answers3