# bounded partial sums of summation of products of sequences related to the harmonic sequence

I would really appreciate any help with regards to the following:

Suppose $\{a_k\}$ and $\{b_k\}$ are sequences that satisfy

• $|a_{k+1}-a_k| \le M/k$ and $|b_{k+1}-b_k| \le M/k$ for all $k$, where $M > 0$
• $0 \le a_k \le N$ and $0 \le b_k \le N$ for all $k$, where $N > 0$.

Does it follow that the sequence of partial sums $S_n := \sum_{k=1}^n (a_{k+1}-a_k) b_k$ is bounded? Or, can a counter-example be constructed?

It seems to me that to make the partial sums $S_n$ unbounded, $a_{k+1}-a_k$ needs to be positive infinitely often, negative infinitely often, and $b_k$ needs to consistently "favor" one of these changes to avoid the cancellations that make $\sum_{k=1}^n(a_{k+1}-a_k)$ bounded.

I tried summation by parts, but this leads to having to analyze the partial sums $\sum_{k=1}^n a_{k+1} (b_{k+1}-b_k)$, which are similar in form to $S_n$.

An interesting and potentially useful fact is that for any positive integer $m$, and $\{a_k\}$ and $\{b_k\}$ that satisfy the above properties, the partial sums $\sum_{k=1}^n (a_{k+1}-a_k) b_k$ are bounded if and only if the partial sums $\sum_{k=1}^n (a_{k+1}-a_k) b_{k+m}$ are bounded.

• Look up summation by parts. – marty cohen Oct 8 '16 at 9:36
• Thanks for the suggestion. I did have a look at that but it led me to a similar question. – gradient23 Oct 8 '16 at 9:46
• $b_{2k} = 0, b_{2k+1} = \frac{1}{2k+2}$, $a_{2k} = 1, a_{2k+1} = 1-\frac{1}{2k+2}$, you get $S_{2n} = \sum_{k=1}^{n} \frac{1}{2k+2}(1+\frac{1}{2k+2})$ – reuns Oct 9 '16 at 14:36
• Thanks @user1952009. With your suggestions for $a_k$ and $b_k$ I am getting $$S_{2n} = \sum_{k=1}^n (a_{2k+1} - a_{2k})b_{2k} + \sum_{k=1}^{n-1} (a_{2(k+1)} - a_{2k+1})b_{2k+1} = \sum_{k=1}^{n-1} \left(\frac{1}{2k+2}\right)^2,$$ which is bounded. – gradient23 Oct 9 '16 at 15:13
• +1 I would like to help good and nice mathematics gradient23. Courage! – user243301 Oct 17 '16 at 16:53

The sequence $(S_n)_{n \in \mathbb{N}}$ need not be bounded. We can arrange it so that $S_n \to +\infty$ (or to $-\infty$), or so that $S_n$ attains arbitrarily large as well as arbitrarily small values, or lots of other behaviours.

It seems to me that to make the partial sums $S_n$ unbounded, $a_{k+1}-a_k$ needs to be positive infinitely often, negative infinitely often, and $b_k$ needs to consistently "favor" one of these changes to avoid the cancellations that make $\sum_{k=1}^n(a_{k+1}-a_k)$ bounded.

Right, if we want $(S_n)$ to converge to $\pm\infty$. If we want $(S_n)$ to oscillate, the preference needs to switch infinitely often, e.g. prefer positive $a_{k+1} - a_k$ until $S_{n_1} > 1$, then prefer negative differences until $S_{n_2} < -2$, then again prefer positive differences until $S_{n_3} > 3$ and so on.

How can we obtain such a behaviour? There are two ways to completely suppress a term $(a_{k+1} - a_k)b_k$, by having $a_{k+1} = a_k$, or by having $b_k = 0$. Now we can create sequences $(a_k)$ and $(b_k)$ satisfying the requirements such that in $\sum_{k = 1}^n (a_{k+1} - a_k)b_k$ all nonzero terms are of the form $(a_{k+1} - a_k)\cdot N$ with $a_k < a_{k+1}$, and each streak of successive nonzero terms adds up to $N^2$. Let

$$n_1 = \min \Biggl\{ n > 1 : \sum_{m = 2}^n \frac{M}{m} \geqslant N\Biggr\},$$

and for $1 \leqslant k < n_1$, set $b_k = N$ and

$$a_k = \sum_{m = 2}^k \frac{M}{m},$$

and $a_{n_1} = N$. This gives us $S_n = a_{n+1}\cdot N$ for $n < n_1$, in particular $S_{n_1-1} = N^2$. Next let

$$n_2 = \min \Biggl\{ n \geqslant n_1 : \sum_{m = n_1}^n \frac{M}{m} \geqslant N\Biggr\}$$

and for $n_1 \leqslant k < n_2$,

$$b_k = N - \sum_{m = n_1}^k \frac{M}{m},$$

and $b_{n_2} = 0$, and $a_k = N$ for $n_1 < k \leqslant n_2 + 1$. Then we have $S_n = N^2$ for $n_1 \leqslant n \leqslant n_2$. Now we let $a_k$ go back to $0$ keeping $b_k = 0$: define

$$n_3 = \min \Biggl\{ n \geqslant n_2 + 2 : \sum_{m = n_2 + 2}^n \frac{M}{m} \geqslant N\Biggr\},$$

and set $b_k = 0$ for $n_2 < k < n_3$, and

$$a_k = N - \sum_{m = n_2+2}^k \frac{M}{m}$$

for $n_2 + 2 \leqslant k < n_3$, and $a_{n_3} = 0$. Then $S_n = N^2$ also for $n_2 < n < n_3$. Next we let $(b_k)$ get back to $N$ while keeping $a_k = 0$: let

$$n_4 = \min \Biggl\{ n \geqslant n_3 : \sum_{m = n_3}^n \frac{M}{m} \geqslant N\Biggr\},$$

set

$$b_k = \sum_{m = n_3}^k \frac{M}{m}$$

for $n_3 \leqslant k < n_4$ and $b_{n_4} = N$, and let $a_k = 0$ for $n_3 < k \leqslant n_4$. This then leaves $S_n = N^2$ for $n_3 \leqslant n < n_4$, and now we can go back to increasing $S_n$, letting $a_k$ climb back to $N$ keeping $b_k = N$ for this streak. Define

$$n_5 = \min \Biggl\{ n > n_4 : \sum_{m = n_4+1}^n \frac{M}{m} \geqslant N\Biggr\},$$

set

$$a_k = \sum_{m = n_4+1}^k \frac{M}{m}$$

for $n_4 < k < n_5$ and $a_{n_5} = N$, and $b_k = N$ for $n_4 < k < n_5$. Continue the construction, defining $n_{j+4}$ analogously to $n_j$ and $a_k, b_k$ analogously on the range determined by $n_{j+3}$ and $n_{j+4}$ to the definitions on the range determined by $n_{j-1}$ and $n_j$. All $n_j$ are well-defined since the harmonic series diverges, and we have $S_{n_{4j-3}-1} = j\cdot N^2$, showing the unboundedness of $(S_n)$. Our construction makes $(S_n)$ monotonic, so $S_n \to +\infty$.

• This is really helpful. Thanks for taking the time and writing the answer. – gradient23 Oct 18 '16 at 7:25