Is there a sequence such that $a_n\to0$, $na_n\to\infty$ and $\left(na_{n+1}-\sum_{k=1}^n a_k\right)$ is convergent?

Is there a positive decreasing sequence $(a_n)_{n\ge 0}$ such that

${\it i.}$ $\lim\limits_{n\to\infty} a_n=0$.

${\it ii.}$ $\lim\limits_{n\to\infty} na_n=\infty$.

${\it iii.}$ there is a real $\ell$ such that $\lim\limits_{n\to\infty} \left(na_{n+1}-\sum_{k=1}^n a_k\right)=\ell.$

Of course without condition ${\it i.}$ constant sequences are non-increasing examples that satisfy the other two conditions, but the requirement that the sequence must be decreasing to zero makes it hard to find!.

Solution 1:

There is no such sequence. Let $s_n=\sum_{k=1}^n a_k$. By (iii) we can write $$ n(s_{n+1}-s_n) - s_n = \ell + b_n = \ell(n+1-n)+ b_n$$ where $b_n\to0$. Then it follows that for all $n\ge1$, $$ \frac{s_{n+1}+\ell}{n+1} = \frac{s_n+\ell}n + \frac{b_n}{n(n+1)} = \frac{s_1+\ell}1 + \sum_{k=1}^n \frac{b_k}{k(k+1)} = A+c_{n+1},$$ where $$ A = s_1+\ell+\sum_{k=1}^\infty \frac{b_k}{k(k+1)}$$ is finite, and $$ c_{n+1} = -\sum_{k=n+1}^\infty \frac{b_k}{k(k+1)},$$ which satisfies $$ |c_n| \le \sup_{k\ge n}|b_k| \sum_{k=n}^\infty \left( \frac1k-\frac{1}{k+1}\right) = \frac1n \sup_{k\ge n}|b_k| $$ Hence $nc_n\to 0$ as $n\to\infty$.

But now it follows $$a_{n+1} = (s_{n+1}+\ell)-(s_n+\ell) = (n+1)(A+c_{n+1})-n(A+c_n) \to A,$$ hence $A=0$ by (i). Then it follows further that $$ s_n + \ell = n c_n \to 0,$$ whence the series $\sum_{k=1}^\infty a_k$ converges to $-\ell$. But this is impossible, since by (ii) we have $a_k\ge1/k$ for $k$ large enough.

Solution 2:

A try:

Suppose that such a sequence exist. Put $\displaystyle na_{n+1}=\sum_{k=1}^n a_k+L+e_n$, with $e_n\to 0$. Replacing $n$ by $n+1$, we get $$(n+1)(a_{n+1}-a_{n+2})=e_n-e_{n+1}$$

As $a_n$ is decreasing, we get that $e_n$ is decreasing too. Now as $e_n\to 0$, we have $e_n\geq 0$, and $\displaystyle a_{n+1}-a_{n+2}=\frac{e_n-e_{n+1}}{n+1} \leq \frac{e_n}{n}-\frac{e_{n+1}}{n+1}$. This show that for $m\geq 2$ we have $$a_{n+1}\leq a_{n+m+2}+\frac{e_n}{n}-\frac{e_{n+m+1}}{n+m+1}\leq a_{n+m+1}+\frac{e_n}{n}$$ If $m\to +\infty$, we get $na_{n+1}\leq e_n$, hence $na_n\to 0$, and condition ii) is not satisfied when conditions (i) and (iii) are.

Solution 3:

Assume $\{a_n\}$ exists.

Let $a_n = f (n)$. By i and ii, we have $f (n) = o (1)$ and $f (n) = \Omega (1/n)$. By Euler-McLaurin summation formula, we have $$\sum_{k = 1}^{n} f (k) = c_1 F (n) + o (F (n))$$ for a constant $c_1$, where $F$ is anti-derivative of $f$, and $F (n) = o (n)$ and $F (n) = \Omega (1)$. Then, by iii, we have

$\begin {eqnarray} n a_{n + 1} - \sum_{k = 1}^{n} a_k & = & n f (n + 1) - \left(c_1 F (n) + o (F (n)\right) \nonumber \\ & = & c_2 F (n) + o (F (n)) - \left(c_1 F (n) + o (F (n))\right) \nonumber \\ & = & (c_1 - c_2) F (n) + o (F (n)) \nonumber \\ & = & O (F (n)) \nonumber \\ & = & \Omega (1), \end {eqnarray}$

but $\Omega (1)_{n \to \infty} = \infty$, a contradiction. Hence, $\{a_n\}$ does not exist.