The sequence $\{a_k\}_{k=1,\cdots,\infty}$ is defined by the following infinite linear recurrence: $$ a_k + \frac{1}{2}a_{k-1} + \frac{1}{3}a_{k-2} + \cdots + \frac{1}{k}a_1 = \frac{1}{k+1} \textrm{ with } a_1=\frac{1}{2}. $$ Is there any way I can prove that $a_k$ is positive for $\forall k$?
I have verified it using MATLAB and it should be true at least until very large $k$. I also tried hard to prove the claim by induction but no luck. I guess some generating function argument may be needed, but I still couldn't figure it out.
I would really appreciate it if anyone could give me a suggestion or reference! Thanks in advance!
Indeed, following the comment by Steven Stadnicki: define $a_0=0$, and define (for $|x|<1$ say) the power series $$ A(x) = \sum_{n=0}^\infty a_k x^k, \quad L(x) = \frac1x \log\frac1{1-x} = \sum_{k=0}^\infty \frac{x^k}{k+1}, \quad\text{and } P(x) = A(x) L(x). $$ Then $P(0)=0$, and for $k\ge1$, the coefficient of $x^k$ in the power series for $P(x)$ is $$ \sum_{j=0}^k \frac1{j+1}a_{k-j} = \frac1{k+1} $$ by assumption, and therefore $P(x) = \sum_{k=1}^\infty \frac{x^k}{k+1} = L(x) - 1$. We conclude that $$ A(x) = \frac{P(x)}{L(x)} = 1-\frac1{L(x)} = 1+\frac x{\log(1-x)}, $$ which can be reality-checked by calcluating the first many $a_k$ directly.
In other words, these $a_k$ are indeed closely related to the "logarithmic numbers" referred to on OEIS. And it seems that it is known that these numbers are positive; the asymptotic formula on the first page of this paper, for example, would imply that positivity.