# If $(n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)$ for all natural $n$, then $a_n$ converges

Per the title, I'm trying to show that if $(n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)$ for all natural $n$, then $a_n$ converges, but I'm getting stuck. Help would be appreciated!

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Let $b_n = a_n - a_{n-1}$. So $b_{n+1} = -\frac{n}{n+1} b_n$, and hence $b_{n+k} = (-1)^k \frac{n}{n+k} b_n$, by induction. With $n=1$, we see that $b_{k+1} = (-1)^k \frac{1}{k} (a_1-a_0)$.
But $$a_n = a_0 + b_1 + b_2 + ... b_n = a_0 + (a_1-a_0) \sum_{k=1}^n \frac{(-1)^k}{k}$$ So the limit converges exactly to $a_0 + (a_1-a_0)\log 2$.
The values of $a_0$ and $a_1$ determine the entire rest of the sequence, and it is easily seen that the defining equation is unaffected by adding a constant to all elements, or multiply all elements by a nonzero constant. These transformations also don't change whether the sequence converges.
Thus, if the sequence is not constant (and therefore certainly convergent), it must be a possibly offset and scaled version of the instance where $a_0=0$ and $a_1=1$. And that sequence is just the partial sums of the alternating harmonic series.