In my circuit theory class, while solving for the steady state voltage of a certain circuit I came across a certain voltage series.
The terms of the series are:
$V_1 = 5 + (0 - 5)e^{-1/k}$
$V_2 = 10 + (V_1 - 10)e^{-1/k}$
$V_3 = 15 + (V_2 - 15)e^{-1/k}$
And so on...
Where $k$ is a constant positive real number.
I think I'll get a recursion like $V_n = 5n + (V_{n-1} - 5n)e^{-1/k}$
I couldn't find the converging value of this series, because first of all I'm not sure how to find the general term of the series. Any idea about how to approach this problem? Will the series even converge?
Finding first few terms, $V_1 = 5 - 5e^{-1/k}$, $V_2 = 5-5e^{-1/k}-5e^{-2/k}$ etc, we come up with the hypotheses $p(n)$ that $$V_n = 5(n-\sum_{r=1}^{n}e^{-1/k})$$
Base case is covered. So for the induction step let this be true for $n = j$. So $V_j = 5(j-\sum_{r=1}^{j}e^{-1/k})$
$$\begin{align}V_{j+1} &= 5(j+1) + (V_j - 5(j+1))e^{-1/k}\\ V_{j+1} &= 5(j+1) + (5j-5\sum_{r=1}^{j}e^{-1/k}- 5j-5))e^{-1/k}\\ &= 5(j+1) -5\sum_{r=1}^{j+1}e^{-1/k} \\ & = 5\left((j+1) - \sum_{r=1}^{j+1}e^{-1/k}\right) \end{align}$$
Thus $p(j) \implies p(j+1)$.