Derive a closed form and find its limit

[lfdahl]

Let $S_n(k)$ be defined by:

$S_n(k) = 1 + 2k+3k^2+...+(n+1)k^n$, where $|k| < 1$ and $n \in \Bbb{N}$.

Derive a closed form for $S_n(k)$ and find the limit: $$\lim_{{n}\to{\infty}}S_n(k)$$.

 

[Opalg]

Let $S_n(k)$ be defined by:

$S_n(k) = 1 + 2k+3k^2+...+(n+1)k^n$, where $|k| < 1$ and $n \in \Bbb{N}$.

Derive a closed form for $S_n(k)$ and find the limit: $$\lim_{{n}\to{\infty}}S_n(k).$$

[sp]Differentiate the geometric series $1 + k + k^2 + \ldots + k^{n+1} = \dfrac{1-k^{n+2}}{1-k}$ with respect to $k$, to get $$1 + 2k + 3k^2 + \ldots + (n+1)k^n = \frac{-(n+2)k^{n+1}(1-k) + 1 - k^{n+2}}{(1-k)^2} = \frac{(n+1)k^{n+2} - (n+2)k^{n+1} + 1}{(1-k)^2} \to \frac1{(1-k)^2}$$ as $n \to \infty$.

[/sp]

 

[lfdahl]

[sp]Differentiate the geometric series $1 + k + k^2 + \ldots + k^{n+1} = \dfrac{1-k^{n+2}}{1-k}$ with respect to $k$, to get $$1 + 2k + 3k^2 + \ldots + (n+1)k^n = \frac{-(n+2)k^{n+1}(1-k) + 1 - k^{n+2}}{(1-k)^2} = \frac{(n+1)k^{n+2} - (n+2)k^{n+1} + 1}{(1-k)^2} \to \frac1{(1-k)^2}$$ as $n \to \infty$.

[/sp]

Thankyou very much, Opalg, for your participation and elegant solution!

 

[MarkFL]

My solution:

Spoiler

We may express $S_n$ in the following difference equation:

$$S_{n}-S_{n-1}=(n+1)k^n$$

The homogeneous solution is:

$$h_n=c_1$$

And the particular solution will take the form:

$$p_n=\left(c_2n+c_3\right)k^n$$

Substitute into our difference equation:

$$\left(c_2n+c_3\right)k^n-\left(c_2(n-1)+c_3\right)k^{n-1}=(n+1)k^n$$

Divide through by $k^{n-1}$:

$$\left(c_2n+c_3\right)k-\left(c_2(n-1)+c_3\right)=(n+1)k$$

Arrange as:

$$c_2(k-1)n+\left(c_3(k-1)+c_2\right)=kn+k$$

Equating like coefficients yields the system:

$$c_2(k-1)=k\implies c_2=-\frac{k}{1-k}$$

$$c_3(k-1)+c_2=c_3(k-1)-\frac{k}{1-k}=k\implies c_3=-\frac{k(2-k)}{(1-k)^2}$$

And thus our particular solution is:

$$p_n=-\left(\frac{k}{1-k}n+\frac{k(2-k)}{(1-k)^2}\right)k^n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}$$

And so by the principle of superposition, we have:

$$S_n=p_n+h_n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}+c_1$$

Using the initial value (extending $n$ to $\mathbb{N_0}$ for simplicity), we find:

$$S_0=-\left(\frac{2-k}{(1-k)^2}\right)k+c_1=1\implies c_1=\frac{1}{(1-k)^2}$$

And so the solution satisfying the given conditions is:

$$S_n=\frac{1}{(1-k)^2}-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}=\frac{1-\left((1-k)n+2-k\right)k^{n+1}}{(1-k)^2}$$

As $$\lim_{n\to\infty}\left(\frac{an+b}{c^n}\right)=0$$ for $1<c$, we see that:

$$S_{\infty}=\lim_{n\to\infty}S_n=\frac{1}{(1-k)^2}$$

 

[lfdahl]

My solution:

Spoiler

We may express $S_n$ in the following difference equation:

$$S_{n}-S_{n-1}=(n+1)k^n$$

The homogeneous solution is:

$$h_n=c_1$$

And the particular solution will take the form:

$$p_n=\left(c_2n+c_3\right)k^n$$

Substitute into our difference equation:

$$\left(c_2n+c_3\right)k^n-\left(c_2(n-1)+c_3\right)k^{n-1}=(n+1)k^n$$

Divide through by $k^{n-1}$:

$$\left(c_2n+c_3\right)k-\left(c_2(n-1)+c_3\right)=(n+1)k$$

Arrange as:

$$c_2(k-1)n+\left(c_3(k-1)+c_2\right)=kn+k$$

Equating like coefficients yields the system:

$$c_2(k-1)=k\implies c_2=-\frac{k}{1-k}$$

$$c_3(k-1)+c_2=c_3(k-1)-\frac{k}{1-k}=k\implies c_3=-\frac{k(2-k)}{(1-k)^2}$$

And thus our particular solution is:

$$p_n=-\left(\frac{k}{1-k}n+\frac{k(2-k)}{(1-k)^2}\right)k^n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}$$

And so by the principle of superposition, we have:

$$S_n=p_n+h_n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}+c_1$$

Using the initial value (extending $n$ to $\mathbb{N_0}$ for simplicity), we find:

$$S_0=-\left(\frac{2-k}{(1-k)^2}\right)k+c_1=1\implies c_1=\frac{1}{(1-k)^2}$$

And so the solution satisfying the given conditions is:

$$S_n=\frac{1}{(1-k)^2}-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}=\frac{1-\left((1-k)n+2-k\right)k^{n+1}}{(1-k)^2}$$

As $$\lim_{n\to\infty}\left(\frac{an+b}{c^n}\right)=0$$ for $1<c$, we see that:

$$S_{\infty}=\lim_{n\to\infty}S_n=\frac{1}{(1-k)^2}$$

Great job, MarkFL! I was not familiar with this approach, so I´ve learned something new! Thanks!(Handshake)

 

[Euge]

Here is my solution.

Spoiler

Note \begin{align}S_n(k) - kS_n(k) &= 1 + (2k - k) + (3k^2 - 2k^2) + \cdots + [(n+1)k^n - nk^n] - (n+1)k^{n+1}\\
(1-k)S_n(k) &= 1 + k + k^2 + \cdots + k^n - (n+1)k^{n+1}\\
(1-k)S_n(k) &= \frac{1-k^{n+1}}{1-k} - (n+1)k^{n+1}\end{align}
Since $k^{n+1}$ and $(n+1)k^{n+1}$ tend to $0$ as $n\to \infty$ (since $\lvert k\rvert < 1$), then setting $L = \lim\limits_{n\to \infty} S_n(k)$, one obtains $(1-k)L = \dfrac{1}{1-k}$, or $L = \dfrac{1}{(1-k)^2}$.

 

[lfdahl]

Here is my solution.

Spoiler

Note \begin{align}S_n(k) - kS_n(k) &= 1 + (2k - k) + (3k^2 - 2k^2) + \cdots + [(n+1)k^n - nk^n] - (n+1)k^{n+1}\\
(1-k)S_n(k) &= 1 + k + k^2 + \cdots + k^n - (n+1)k^{n+1}\\
(1-k)S_n(k) &= \frac{1-k^{n+1}}{1-k} - (n+1)k^{n+1}\end{align}
Since $k^{n+1}$ and $(n+1)k^{n+1}$ tend to $0$ as $n\to \infty$ (since $\lvert k\rvert < 1$), then setting $L = \lim\limits_{n\to \infty} S_n(k)$, one obtains $(1-k)L = \dfrac{1}{1-k}$, or $L = \dfrac{1}{(1-k)^2}$.

What a nice approach, Euge! Thankyou very much for your participation!(Yes)